ukmuiik

Question

How can I evaluate
$$sum_n=1^inftyfrac2n3^n+1$$
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before so I feel that there must be a simpler method.

In general, how can I evaluate $$sum_n=0^infty (n+1)x^n?$$

It is the other way around. The linked question might be duplicate of this one. — Sep 21 '13 at 17:34
I believe this is an arithmo-geometric series. You can find information here: artofproblemsolving.com/wiki/â€¦ — Jul 7 at 19:11

Parcly Taxel 33.5k136588 · Accepted Answer · 2017-09-24 12:07:56Z

No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: $$S_m=sum_n=1^mnr^n.$$

Notice that
beginalign*
S_m-rS_m & = -mr^m+1+sum_n=1^mr^n\
& = -mr^m+1+fracr-r^m+11-r \
& =fracmr^m+2-(m+1)r^m+1+r1-r.
endalign*

Hence
$$S_m = fracmr^m+2-(m+1)r^m+1+r(1-r)^2.$$

This equality holds for any $r$, but in your case we have $r=frac13$ and a factor of $frac23$ in front of the sum. That is
beginalign*
sum_n=1^inftyfrac2n3^n+1
& = frac23lim_mrightarrowinftyfracmleft(frac13right)^m+2-(m+1)left(frac13right)^m+1+left(frac13right)left(1-left(frac13right)right)^2 \
& =frac23fracleft(frac13right)left(frac23right)^2 \
& =frac12.
endalign*

Added note:

We can define $$S_m^k(r) = sum_n=1^m n^k r^n.$$ Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$. We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$. This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for.

This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_m^k(r)$ in general for any $k$? It turns out we can, and the formula is similar to the formula for $sum_n=1^m n^k$, and involves the Bernoulli numbers. In particular, the denominator is $(1-r)^k+1$.

@Eric How do you make the transformation $sum_n=1^m=r-r^m+1 over 1-r$? Secondly at this step you can substitute the series with this explicit formula as the series converges (obviously because it is finite). If the series was infinite you couldn't have done that (unless $|r| lt 1$) as it would diverge. However later you apply this formula to an infinite series $sum_n=1^infty2n over 3^n+1$. Could you explain why you consider it to be suitable for an infinite series, albeit it was initially brought out for finite series? — Jun 5 '14 at 16:38
Small nitpick: "equality holds for any $r$" should be "equality holds for any $rneq1$" — Feb 25 '17 at 5:27

Micah 27.8k135899 · Answer 2 · 2012-10-29 22:43:49Z

up vote
198
down vote

If you want a solution that doesn't require derivatives or integrals, notice that
begineqnarray
1+2x+3x^2+4x^3+dots = 1 + x + x^2 + x^3 + dots \
+ x + x^2+ x^3 + dots\
+ x^2 + x^3 + dots \
+x^3 + dots \
+ dots \
=1 + x + x^2 + x^3+dots \
+x(1+x+x^2+dots) \
+x^2(1+x+dots)\
+x^3(1+dots)\
+dots \
=(1+x+x^2+x^3+dots)^2=frac1(1-x)^2
endeqnarray

answered Oct 29 '12 at 22:43

Micah

27.8k135899

16

Your solution has a large gap. You will find it difficult to prove that the series converges without using as much technical machinery as the solution that goes through the finite sum before taking limits.
â€“Â user21820
Aug 16 '15 at 5:39

2

@user21820: The proof as it stands (replacing the ellipses by a precise description of the general terms they stand for) is perfectly valid for if expressions are interpreted as formal power series in$~x$, in other words it shows that $sum_ngeq0(n+1)X^n=(1-X)^-2$ in $Bbb Z[[X]]$. With this established (or actually independently of it) is it very easy to show that the power series in both members have radius of convergence$~1$ (it suffices to observe that the coefficients increase, but less than exponentially), obtaining an identity of convergent power series within that radius.
â€“Â Marc van Leeuwen
Feb 24 '17 at 5:50

@MarcvanLeeuwen: You say "very easy", but it is still much easier to just truncate at the finite terms because the remainder term goes to zero by elementary means. That is why I said very precisely "without using as much technical machinery as ...", which is practically none. Micah's answer as stated is extremely misleading to students, for the same reason that most people can't see the flaw in the proof of $1+2+3+cdots = -frac112$? Nevertheless, if one wants to develop the general tool of generating functions for combinatorics, then yes we should develop formal power series.
â€“Â user21820
Feb 24 '17 at 5:55

2

@MarcvanLeeuwen: Besides, I am appalled at the general lack of rigour in half the answers here, especially since the question itself clearly states that Wolfram Alpha cites convergence tests and asks for a simpler way, so anything that requires convergence tests is not simpler.
â€“Â user21820
Feb 24 '17 at 6:06

5

@user21820: I don't understand what you mean. I don't see how the proof using truncation would go precisely, nor how it would easy because the crucial factoring of the sum as a product of two (equal) sums is only valid for the infinite sum. Your original comment does not sound as if it want to say (rather vacuously) "without using at least $0$ effort" either. Remains that you worry about misleading your students; I disagree. There is nothing wrong with the approach of proving something for formal power series first, then considering convergence afterwards. We should teach our students that
â€“Â Marc van Leeuwen
Feb 24 '17 at 6:08

Â |Â
show 1 more comment

Jonas Meyer 39k6140246 · Answer 3 · 2011-04-03 22:25:54Z

As indicated in other answers, you can reduce this to summing $displaystylesum_n=1^infty na^n$ with $|a|<1$ (by pulling out the constant $frac23$ and rewriting with $a=frac13$). This in turn can be reduced to summing geometric series by rearranging and factoring. Note that, assuming everything converges nicely (which it does):

$beginmatrix
&a & + & 2a^2 & + & 3a^3 &+& 4a^4 &+& cdots\
=&a &+& a^2 &+& a^3 &+& a^4 &+& cdots\
+& & & a^2 &+& a^3 &+& a^4 &+& cdots\
+& & & & & a^3 &+& a^4 &+& cdots\
+& & & & & & & a^4 &+& cdots\
+& & & & & & & & & vdots
endmatrix$

Factoring out the lowest power of $a$ in each row yields

$beginalign*
sum_n=1^infty na^n
&= a(1+a^2+a^3+cdots)\
&+ a^2(1+a^2+a^3+cdots)\
&+ a^3(1+a^2+a^3+cdots)\
&+ a^4(1+a^2+a^3+cdots)\
&vdots
endalign*$

Each row in the last expression has the common factor $a(1+a+a^2+a^3+cdots)$, and factoring this out yields

$beginalign*sum_n=1^infty na^n
&=a(1+a+a^2+a^3+cdots)(1+a+a^2+a^3+cdots)\
&=a(1+a+a^2+a^3+cdots)^2.endalign*$

Now you can finish by summing the geometric series.

Eric Naslund's answer was posted while I was writing, but I thought that this alternative approach might be worth posting. I also want to mention that in general one should be careful about rearranging series as though they were finite sums. To be more formal, some of the steps above would require justification based on absolute convergence.

DarÃo A. GutiÃ©rrez 2,40721129 · Answer 4 · 2016-09-22 11:29:17Z

Factor out the $frac23$. Then write $$sum_n=1^infty fracn3^n = sum_n=1^infty frac13^n + sum_n=2^infty frac13^n + sum_n=3^infty frac13^n + cdots$$

It is easy to show that $$sum_n=m^infty frac13^n = frac32 left(frac13 right)^m$$
and so
$$sum_n=1^infty fracn3^n = frac32 sum_n=1^infty left( frac13 right)^n $$
which you can sum. Don't forget to put the $frac23$ back in.

score 82 · Answer 5 · 2015-01-28 16:19:41Z

My favorite proof of this is in this paper of Roger B. Nelsen

I also have the following method for $sum_n=1^infty nover 2^n-1$ (one can use a similar method for $sum_n=1^infty nover3^n$):

We first show that $sumlimits_n=7^infty nover 2^n-1 =1over4$.

We start with a rectangle of width 1 and height $1/4$. Divide this into eights:
enter image description here

Now divide each eighth-rectangle above in half and take 7 of them. This gives $A_1=7over 2^6$.

enter image description here
There are $2cdot8-7=9$ boxes left over, each having area $2^-6$.

Divide each remaining $16^rm th$-rectangle in half and take 8 of them. This gives $A_2=7over 2^6+8over 2^7$.

enter image description here
There are $2cdot9-8=10$ boxes left over, each having area $2^-7$.

Divide each remaining $32^rm nd$-rectangle in half and take 9 of them. This gives $A_3=7over 2^6+8over 2^7+9over 2^8$.

enter image description here
There are $2cdot10-9=11$ boxes left over, each having area $2^-8$.

Divide each remaining $64^rm th$-rectangle in half and take 10 of them. This gives $A_4=7over 2^6+8over 2^7+9over 2^8+10over2^9$.

enter image description here
There are $2cdot11-9=12$ boxes left over, each having area $2^-9$.

At each stage, we double the number of remaining boxes, keeping the same leftover area, and take approximately
half of them to form the next term of the series.

At the $n^rm th$ stage, we have $$A_n= 7over 2^6+8over 2^7+cdots+6+nover2^5+n,$$

with leftover area $$ 2(n+7)-(n+6)over 2^n+5.$$

It follows that,
$$
7over2^6+8over2^7+9over2^8+cdots= 1over4.
$$
Consequently,
$$
sum_n=1^inftynover 2^n-1= sum_n=1^6 nover 2^n-1 +sum_n=7^inftynover 2^n-1 =15over 4+1over4=4.
$$

You can also "Fubini" this (I think this is what Jonas is doing).

Mitch 5,9062456 · Answer 6 · 2011-04-03 22:51:55Z

up vote
67
down vote

Hints

You know (don't you?) the formula for $S(a) = sum_n=0^infty a^n$ for $|a| < 1$

Take the derivative (with respect to $a$) of both sides to obtain a formula for $sum_n=1^infty n a^n$

Show that your series can be put in that form.

edited Apr 3 '11 at 22:51

Mitch

5,9062456

answered Apr 3 '11 at 21:54

leonbloy

38k644104

Thanks for answering. 1) No I don't know that formula 2) Can you explain by what you mean by derive both sides?
â€“Â Backus
Apr 3 '11 at 21:59

1

1. See here: en.wikipedia.org/wiki/Geometric_series 2) Derivation (or differentiation) is a mathematical operation (well, more than that) that is taught in Calculus - if you don't know about it, forget about my answer (and xen0m's). Perhaps you can solve your problem without knowing Calculus, by other methods... but normally you first learn calculus, then solve series.
â€“Â leonbloy
Apr 3 '11 at 22:05

add a commentÂ |Â

score 65 · Answer 7 · 2015-11-20 18:08:32Z

Note that $int 1 + 2x + 3x^2 + cdots , dx = x + x^2 + x^3 + cdots + textconst$, i.e., a geometric series, which converges to $x/(1 - x)$ if $|x| < 1$. Therefore,
$$fracddx left(fracx1 - xright) = frac(1 - x)(1) - x(-1)(1 - x)^2 = frac1(1 - x)^2,$$
that is,
$$1 + 2x + 3x^2 + cdots = frac1(1 - x)^2.$$

Another proof: Let $S = 1 + 2x + 3x^2 + cdots$ with $|x| < 1$. Then
$$xS = x + 2x^2 + 3x^3 + cdots$$
so
$$S - xS = (1- x)S = 1 + x + x^2 + cdots = frac11- x.$$
Therefore: $S = (1 - x)^-2$.

+1, but wouldn't it have been simpler to say that it was the derivative of $1+x+x^2+...$, converging to $frac11-x$? — Oct 29 '12 at 22:46
I decided to start with what was given, so it is easier for the OP to see. — Oct 29 '12 at 22:48
What I meant is that you chose $c=0$ while $c=1$ is a more well known series and is easier to take the derivative of. — Oct 29 '12 at 23:23
We could notice that the given series is converging to $frac11-x-1$ and take the derivative of that. — Nov 11 '14 at 21:03
@inkievoyd That is exactly what I did. Note that $(1 - x)^-1 - 1 = x(1 - x)^-1$. — Nov 12 '14 at 11:47

xen 2,9841523 · Answer 8 · 2011-04-03 21:55:25Z

You can find by differentiation. Just notice that $(x^n)' = nx^n-1$. By the theory of power series we obtain (by uniform convergence on any compact subset of $(-1,1)$) that
$$
left(sum_n=1^infty x^nright)' = sum_n=1^infty (x^n)' = sum_n=1^infty n x^n-1.
$$
The sum on the left hand side is equal to $left(fracx1-xright)'$. You need to notice that your sum can be written in a similar way as $sum_n=1^infty nx^n-1$.

Thank you for helping, but I have never learned differentiation. — Apr 3 '11 at 21:58
The sum $sum x^n$ is equal to $dfrac11-x$ and therefore $sum nx^n=xleft(frac11-xright)'$ gives the correct result instead of $left(fracx1-xright)'$. — Mar 9 '16 at 17:23

1233dfv 4,0401225 · Answer 9 · 2013-10-25 20:54:57Z

Consider the generating function $$g(x)=sum_n=0^inftyn+k-1choose nx^n=1over (1-x)^k.$$ If we let $k=2$, then $$sum_n=0^inftyn+1choose nx^n=1over (1-x)^2.$$ Since $n+1choose n=(n+1)$ we can conclude that $$sum_n=0^infty(n+1)x^n=1over (1-x)^2.$$

score 43 · Answer 10 · 2013-11-01 16:09:40Z

Let be $$S_n(z)=sum_j=1^+inftyj^nz^jquadtextfor zinBbbC, |z|<1, n=0, 1, 2, ldots $$
It's easy to prove that for $zinBbbC, |z|<1$, the sums $S_n(z)$ satisfy the auto-convolutional recurrence relation
$$
S_n+1(z)=S_n(z)+sum_k=0^nbinomnk S_k(z)S_n-k(z)qquad n=0, 1, 2, ldots
$$
Infact, performing the change index $q=j-i$ and using binomial theorem, we have
$$
beginalign
S_n+1(z)&=sum_j=1^+inftyj^n+1z^j=sum_j=1^+inftyj^nz^j+sum_i=1^+inftysum_j=i+1^+inftyj^nz^j\
&=S_n(z)+sum_i=1^+inftysum_q=1^+infty(i+q)^nz^i+q\
&=S_n(z)+sum_i=1^+inftysum_q=1^+inftysum_k=0^nbinomnki^kq^n-kz^iz^q\
&=S_n(z)+sum_k=0^nbinomnksum_i=1^+inftyi^kz^isum_q=1^+inftyq^n-kz^q\
&=S_n(z)+sum_k=0^nbinomnk S_k(z)S_n-k(z)
endalign
$$

For $n = 0$ the sum $S_0(z)$ is the sum of geometric progression
$$
S_0(z)=sum_j=1^+inftyz^j=fracz1-z
$$
Using the recurrence we find
$$
beginalign
S_1(z)&=S_0(z)+S_0^2(z)=fracz(1-z)^2\
S_2(z)&=S_1(z)+2S_0(z)S_1(z)=fracz^2+z(1-z)^3\
S_3(z)&=S_2(z)+2S_0(z)S_2(z)+S_1^2(z)=fracz^3+4z^2+z(1-z)^4
endalign
$$
and so on.

Using the founded results, for $a, b, z inBbbC, zneq 0,|z|<1$, putting $$sigma(z;a,b)=sum_j=0^+infty(a+bj) z^j$$ one has
$$
sigma(z;a,b)=sum_j=0^+infty(a+bj) z^j=a[1+S_0(z)]+bS_1(z)=fraca+(b-a)z(1-z)^2
$$

So the required sum is
$$
sum_n=0^+infty(n+1) x^n=sigma(x;1,1)=frac1(1-z)^2
$$
and
$$
sum_n=1^+inftyfrac2n3^n+1=frac23^2sigmaleft(frac13;1,1right)=frac12
$$

Note In alternative to the auto-convolution relation we can use another useful recursive relation for $zinBbbC, |z|<1$, that is the linear recurrence
$$
S_n(z)=fracz1-zleft[1+sum_k=0^n-1binomnk S_k(z)right]qquad n=1, 2, ldots
$$

score 41 · Answer 11 · 2014-09-05 12:30:37Z

Note that $n+1$ is the number ways to choose $n$ items of $2$ types (repetitions allowed but order is ignored), so that $n+1=left(!binom2n!right)=(-1)^nbinom-2n$. (This uses the notation $left(!binom mn!right)$ for the number of ways to choose $n$ items of $m$ types with repetition, a number equal to $binomm+n-1n=(-1)^nbinom-mn$ by the usual definiton of binomial coefficients with general upper index.) Now recognise the binomial formula for exponent $-2$ in
$$
sum_ngeq0(n+1)x^n=sum_ngeq0(-1)^ntbinom-2nx^n
=sum_ngeq0tbinom-2n(-x)^n=(1-x)^-2.
$$
This is valid as formal power series in$~x$, and also gives an identity for convergent power series whenever $|x|<1$.

There is a nice graphic way to understand this identity. The terms of the square of the formal power series $frac11-x=sum_igeq0x^i$ can be arranged into an infinite matrix, with at position $(i,j)$ (with $i,jgeq0$) the term$~x^i+j$ . Now for given $n$ the terms $x^n$ occur on the $n+1$ positions with $i+j=n$ (an anti-diagonal) and grouping like terms results in the series $sum_ngeq0(n+1)x^n$.

And notice that I didn't comment on your answer because you explicitly stated "formal power series" and also precisely stated the convergence radius. Now I do not believe this actually answers the question, for the reason I already gave you, namely that your answer requires tools that are not simpler than the convergence test used by WA, as requested in the question. But I have nothing wrong with your answer, since it is not mathematically incorrect or misleading. — Feb 24 '17 at 6:25

MathFacts 2,9891424 · Answer 12 · 2014-07-27 07:32:46Z

In fact,
$$
sum_n=0^+infty(n+1)x^n = sum_n=0^+inftyfracddx(x^n+1)= fracddxsum_n=0^+inftyx^n+1 = fracddxbiggl(fracx1 - xbiggr) = frac1(1 - x)^2
$$
For $x = frac13$, we have
$$
frac94 =sum_n=0^+infty(n+1)frac13^n = sum_m=1^+inftymfrac13^m-1 quad Rightarrow quad sum_m=1^+inftyfracm3^m = frac34
$$

Marc van Leeuwen 84.8k499212 · Answer 13 · 2014-09-05 12:33:25Z

I assume that the $|x|$ to be less than $1$. Now, consider,
$f(x)=sum_n=0^n=infty x^n+1$

This will converge only if $|x|<1$. Now, interesting thing here is, this is a geometric progression. The $f(x)=x/(1-x)$.

$f'(x)$ is the series you are interested in, right? Differentiate $x/(1-x)$ and you have your expression!

John Joy 5,88511526 · Answer 14 · 2014-12-11 17:17:27Z

I first encountered this sum with the following problem:

Evaluate
$$bigg(frac12bigg)^dfrac13bigg(frac14bigg)^dfrac19bigg(frac18bigg)^dfrac127bigg(frac116bigg)^dfrac181dots$$

Which , of course simplified to
$$bigg(frac12bigg)^dfrac13^1+dfrac23^2+dfrac33^3+dfrac43^4+dots=bigg(frac12bigg)^S$$

Getting back to your problem, now
$$sum_n=1^infty frac2n3^n+1=frac23sum_n=1^infty fracn3^n=frac23S$$
Using a method similar to deriving geometric series suppose that
$$S_k = sum_n=1^k fracn3^n$$
Then we have
$$beginarraylll
3S_k-S_k &=& 1+frac23^1+frac33^2+frac43^3+dots+frack3^k-1\
&&-frac13^1-frac23^2-frac33^3dots-frack-13^k-1-frack3^k\
2S_k&=&bigg(1+frac2-13^1+frac3-23^2+frac4-33^3+dots+frack-(k-1)3^k-1bigg)-frack3^k\
&=&frac1-(frac13)^k1-frac13 - frack3^k\
2S&=&lim_ktoinfty frac1-(frac13)^k1-frac13 - frack3^k\
2S&=&frac11-frac13=frac32\
frac23S&=&frac12\
endarray$$

by a similar method one can show, that if the series converges, that
$$sum_n=0^infty (n+1)x^n = frac1(1-x)^2$$

Olivier Oloa 106k17173292 · Answer 15 · 2016-03-20 17:35:21Z

To avoid differentiating an infinite sum.

We start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=frac1-x^n+11-x, quad |x|<1. tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^n-1=frac1-x^n+1(1-x)^2+frac-(n+1)x^n1-x, quad |x|<1, tag2
$$ and by making $n to +infty$ in $(2)$, using $|x|<1$, gives

$$
sum_n=0^infty(n+1)x^n=frac1(1-x)^2. tag3
$$

Rodrigo de Azevedo 12.6k41751 · Answer 16 · 2018-03-02 12:10:58Z

One method of evaluating $sum_n=0^infty(1+n)x^n$ can be like this, we take the generating function $$f = sum_n=0^infty x^n $$ then $$sum_n=0^infty (n+1)x^n = (xD + 1) f $$ $$ fracx(1-x)^2 + frac11-x = frac1(1-x)^2$$ where $D$ means differentiation w.r.t. $x$.

score 12 · Answer 17 · 2017-04-08 15:07:42Z

No one like finite calculus notation? Unbelievable :(

I must add an answer in the form of finite calculus. You can read about this topic in the book Concrete Mathematics of Graham and Knuth, or this paper.

Finite calculus is analogous to the normal (infinitesimal) calculus where we use instead "discrete derivatives" and "discrete integrals" (actually just summations), and we can perform definite or indefinite sums in analogy to definite or indefinite integrals.

Analogously to the standard derivative the discrete derivative and the discrete (indefinite) integral can be written as

$$Delta f(k):=f(k+1)-f(k),quadquad sum f(k)delta k=F(k)+Ctag1$$

for some $1$-periodic function $C$, and where we have too that

$$sum_k=a^bf(k)=sumnolimits_a^b+1f(k)delta ktag2$$

And we have the summation by parts formula with this symbology represented by

$$sum f(k)[Delta g(k)]delta k=f(k)g(k)-sum mathrm [E g(k)]f(k)delta ktag3$$

where $mathrm E$ is the shift operator and is defined as $mathrm E f(k):=f(k+1)$. By last, before to answer the question, it is not hard to check that

$$Delta x^k=x^k(x-1),quadsum x^kdelta k=x^k(x-1)^-1+C\Delta (k+w)=1,quad sum (k+w)delta k=frac12 (k+w-1)(k+w)+Ctag4$$

Hence, using the above formulas, we have that

$$beginalignsum_k=0^infty (k+1)x^k&=sumnolimits_0^infty (k+1)x^kdelta k\&=(k+1)x^k(x-1)^-1big|_0^infty-sumnolimits_0^infty x^k+1(x-1)^-1delta k\&=[(k+1)x^k(x-1)^-1-x^k+1(x-1)^-2]big|_0^inftyendalign$$

Then the above is finite when $|x|<1$, in this case we have that

$$sum_k=0^infty (k+1)x^k=-frac1x-1+fracx(x-1)^2=frac1(x-1)^2$$

DarÃo A. GutiÃ©rrez 2,40721129 · Answer 18 · 2018-02-24 18:48:38Z

beginalign
sum_n=0^infty (n+1)x^n &= sum_n=1^infty nx^n-1 =fracddxleft( sum_n=0^infty x^nright) =fracddxleft(frac11-xright)=frac1(1-x)^2
endalign
beginalign
tag*$Box$
endalign

robjohnâ™¦ 258k26297612 · Answer 19 · 2018-04-16 12:10:52Z

up vote
2
down vote

Solving $(an+b)-(a(n+1)+b)x=n+1$ for all $n$ gives $a=frac11-x$ and $b=frac1(1-x)^2$. Therefore,
$$
(n+1)x^n=left(frac1(1-x)^2+fracn1-xright)x^n-left(frac1(1-x)^2+fracn+11-xright)x^n+1tag1
$$
Using $(1)$ and telescoping series, we get
$$
sum_k=0^n-1(k+1)x^k=frac1(1-x)^2-left(frac1(1-x)^2+fracn1-xright)x^ntag2
$$
If $|x|lt1$, then we get
$$
sum_k=0^infty(k+1)x^k=frac1(1-x)^2tag3
$$

answered Apr 16 at 12:10

robjohnâ™¦

258k26297612

Hello robjohn, i need your help on that technique of generating valid telescoping series if you allow this ofc.
â€“Â Abr001am
Jun 6 at 20:29

@Abr001am: what is your question?
â€“Â robjohnâ™¦
Jun 7 at 5:54

robjohn, you used this generating system of two equations that got you landed on a valid interleaving series, i tried as much with other series of different forms as often i return back from the starting point, feel like parcoursing a void circle, how can you tell if a series is reducible to a series of that sort ?
â€“Â Abr001am
Jun 9 at 15:42

@Abr001am: I think it is usually possible, but it may be as hard to find the telescoping terms as it is to find the closed form for the sum.
â€“Â robjohnâ™¦
Jun 9 at 17:49

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Parcly Taxel 33.5k136588 · Accepted Answer · 2017-09-24 12:07:56Z

No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: $$S_m=sum_n=1^mnr^n.$$

Notice that
beginalign*
S_m-rS_m & = -mr^m+1+sum_n=1^mr^n\
& = -mr^m+1+fracr-r^m+11-r \
& =fracmr^m+2-(m+1)r^m+1+r1-r.
endalign*

Hence
$$S_m = fracmr^m+2-(m+1)r^m+1+r(1-r)^2.$$

This equality holds for any $r$, but in your case we have $r=frac13$ and a factor of $frac23$ in front of the sum. That is
beginalign*
sum_n=1^inftyfrac2n3^n+1
& = frac23lim_mrightarrowinftyfracmleft(frac13right)^m+2-(m+1)left(frac13right)^m+1+left(frac13right)left(1-left(frac13right)right)^2 \
& =frac23fracleft(frac13right)left(frac23right)^2 \
& =frac12.
endalign*

Added note:

We can define $$S_m^k(r) = sum_n=1^m n^k r^n.$$ Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$. We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$. This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for.

This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_m^k(r)$ in general for any $k$? It turns out we can, and the formula is similar to the formula for $sum_n=1^m n^k$, and involves the Bernoulli numbers. In particular, the denominator is $(1-r)^k+1$.

@Eric How do you make the transformation $sum_n=1^m=r-r^m+1 over 1-r$? Secondly at this step you can substitute the series with this explicit formula as the series converges (obviously because it is finite). If the series was infinite you couldn't have done that (unless $|r| lt 1$) as it would diverge. However later you apply this formula to an infinite series $sum_n=1^infty2n over 3^n+1$. Could you explain why you consider it to be suitable for an infinite series, albeit it was initially brought out for finite series? — Jun 5 '14 at 16:38
Small nitpick: "equality holds for any $r$" should be "equality holds for any $rneq1$" — Feb 25 '17 at 5:27

Micah 27.8k135899 · Answer 21 · 2012-10-29 22:43:49Z

up vote
198
down vote

If you want a solution that doesn't require derivatives or integrals, notice that
begineqnarray
1+2x+3x^2+4x^3+dots = 1 + x + x^2 + x^3 + dots \
+ x + x^2+ x^3 + dots\
+ x^2 + x^3 + dots \
+x^3 + dots \
+ dots \
=1 + x + x^2 + x^3+dots \
+x(1+x+x^2+dots) \
+x^2(1+x+dots)\
+x^3(1+dots)\
+dots \
=(1+x+x^2+x^3+dots)^2=frac1(1-x)^2
endeqnarray

answered Oct 29 '12 at 22:43

Micah

27.8k135899

16

Your solution has a large gap. You will find it difficult to prove that the series converges without using as much technical machinery as the solution that goes through the finite sum before taking limits.
â€“Â user21820
Aug 16 '15 at 5:39

2

@user21820: The proof as it stands (replacing the ellipses by a precise description of the general terms they stand for) is perfectly valid for if expressions are interpreted as formal power series in$~x$, in other words it shows that $sum_ngeq0(n+1)X^n=(1-X)^-2$ in $Bbb Z[[X]]$. With this established (or actually independently of it) is it very easy to show that the power series in both members have radius of convergence$~1$ (it suffices to observe that the coefficients increase, but less than exponentially), obtaining an identity of convergent power series within that radius.
â€“Â Marc van Leeuwen
Feb 24 '17 at 5:50

@MarcvanLeeuwen: You say "very easy", but it is still much easier to just truncate at the finite terms because the remainder term goes to zero by elementary means. That is why I said very precisely "without using as much technical machinery as ...", which is practically none. Micah's answer as stated is extremely misleading to students, for the same reason that most people can't see the flaw in the proof of $1+2+3+cdots = -frac112$? Nevertheless, if one wants to develop the general tool of generating functions for combinatorics, then yes we should develop formal power series.
â€“Â user21820
Feb 24 '17 at 5:55

2

@MarcvanLeeuwen: Besides, I am appalled at the general lack of rigour in half the answers here, especially since the question itself clearly states that Wolfram Alpha cites convergence tests and asks for a simpler way, so anything that requires convergence tests is not simpler.
â€“Â user21820
Feb 24 '17 at 6:06

5

@user21820: I don't understand what you mean. I don't see how the proof using truncation would go precisely, nor how it would easy because the crucial factoring of the sum as a product of two (equal) sums is only valid for the infinite sum. Your original comment does not sound as if it want to say (rather vacuously) "without using at least $0$ effort" either. Remains that you worry about misleading your students; I disagree. There is nothing wrong with the approach of proving something for formal power series first, then considering convergence afterwards. We should teach our students that
â€“Â Marc van Leeuwen
Feb 24 '17 at 6:08

Â |Â
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Jonas Meyer 39k6140246 · Answer 22 · 2011-04-03 22:25:54Z

As indicated in other answers, you can reduce this to summing $displaystylesum_n=1^infty na^n$ with $|a|<1$ (by pulling out the constant $frac23$ and rewriting with $a=frac13$). This in turn can be reduced to summing geometric series by rearranging and factoring. Note that, assuming everything converges nicely (which it does):

$beginmatrix
&a & + & 2a^2 & + & 3a^3 &+& 4a^4 &+& cdots\
=&a &+& a^2 &+& a^3 &+& a^4 &+& cdots\
+& & & a^2 &+& a^3 &+& a^4 &+& cdots\
+& & & & & a^3 &+& a^4 &+& cdots\
+& & & & & & & a^4 &+& cdots\
+& & & & & & & & & vdots
endmatrix$

Factoring out the lowest power of $a$ in each row yields

$beginalign*
sum_n=1^infty na^n
&= a(1+a^2+a^3+cdots)\
&+ a^2(1+a^2+a^3+cdots)\
&+ a^3(1+a^2+a^3+cdots)\
&+ a^4(1+a^2+a^3+cdots)\
&vdots
endalign*$

Each row in the last expression has the common factor $a(1+a+a^2+a^3+cdots)$, and factoring this out yields

$beginalign*sum_n=1^infty na^n
&=a(1+a+a^2+a^3+cdots)(1+a+a^2+a^3+cdots)\
&=a(1+a+a^2+a^3+cdots)^2.endalign*$

Now you can finish by summing the geometric series.

Eric Naslund's answer was posted while I was writing, but I thought that this alternative approach might be worth posting. I also want to mention that in general one should be careful about rearranging series as though they were finite sums. To be more formal, some of the steps above would require justification based on absolute convergence.

DarÃo A. GutiÃ©rrez 2,40721129 · Answer 23 · 2016-09-22 11:29:17Z

Factor out the $frac23$. Then write $$sum_n=1^infty fracn3^n = sum_n=1^infty frac13^n + sum_n=2^infty frac13^n + sum_n=3^infty frac13^n + cdots$$

It is easy to show that $$sum_n=m^infty frac13^n = frac32 left(frac13 right)^m$$
and so
$$sum_n=1^infty fracn3^n = frac32 sum_n=1^infty left( frac13 right)^n $$
which you can sum. Don't forget to put the $frac23$ back in.

score 82 · Answer 24 · 2015-01-28 16:19:41Z