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Evaluating $frac1^22^1+frac3^22^2+frac5^22^3+cdots$ using sigma notation
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This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$frac1^22^1+frac3^22^2+frac5^22^3+cdots+frac(2r +1)^22^r+cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$frac4n(n+1)(2n+1)6 + n + frac4(ncdot n+1)2$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$
sequences-and-series
edited Jul 27 at 18:49
Blue
43.6k868141
asked Jul 27 at 18:03
jame samajoe
288
 |Â
show 6 more comments
up vote
3
down vote
favorite
This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$frac1^22^1+frac3^22^2+frac5^22^3+cdots+frac(2r +1)^22^r+cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$frac4n(n+1)(2n+1)6 + n + frac4(ncdot n+1)2$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$
sequences-and-series
edited Jul 27 at 18:49
Blue
43.6k868141
asked Jul 27 at 18:03
jame samajoe
288
everything is getting messed up as i m applying infinity while doing summation what to do where am i wrong
– jame samajoe
Jul 27 at 18:12
You... forgot the factor $1/2^r$ in your 3 resulting summations.
– Clement C.
Jul 27 at 18:32
@ClementC. can u explain what u want to say
– jame samajoe
Jul 27 at 18:35
You computed $$sum_r=1^n 4r^2 + sum_r=1^n 1+ sum_r=1^n 4r$$ instead of $$sum_r=1^n frac4r^22^r + sum_r=1^n frac12^r+ sum_r=1^n frac4r2^r$$
– Clement C.
Jul 27 at 18:36
1
@jamesamajoe Not that the r-th summand is $$frac(2r colorred-1)^22^r$$ Y
– callculus
Jul 27 at 18:54
 |Â
show 6 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$frac1^22^1+frac3^22^2+frac5^22^3+cdots+frac(2r +1)^22^r+cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$frac4n(n+1)(2n+1)6 + n + frac4(ncdot n+1)2$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$
sequences-and-series
edited Jul 27 at 18:49
Blue
43.6k868141
asked Jul 27 at 18:03
jame samajoe
288
This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$frac1^22^1+frac3^22^2+frac5^22^3+cdots+frac(2r +1)^22^r+cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$frac4n(n+1)(2n+1)6 + n + frac4(ncdot n+1)2$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$
sequences-and-series
edited Jul 27 at 18:49
Blue
43.6k868141
asked Jul 27 at 18:03
jame samajoe
288
edited Jul 27 at 18:49
Blue
43.6k868141
edited Jul 27 at 18:49
Blue
43.6k868141
edited Jul 27 at 18:49
Blue
43.6k868141
43.6k868141
asked Jul 27 at 18:03
jame samajoe
288
asked Jul 27 at 18:03
jame samajoe
288
asked Jul 27 at 18:03
jame samajoe
288
288
everything is getting messed up as i m applying infinity while doing summation what to do where am i wrong
– jame samajoe
Jul 27 at 18:12
You... forgot the factor $1/2^r$ in your 3 resulting summations.
– Clement C.
Jul 27 at 18:32
@ClementC. can u explain what u want to say
– jame samajoe
Jul 27 at 18:35
You computed $$sum_r=1^n 4r^2 + sum_r=1^n 1+ sum_r=1^n 4r$$ instead of $$sum_r=1^n frac4r^22^r + sum_r=1^n frac12^r+ sum_r=1^n frac4r2^r$$
– Clement C.
Jul 27 at 18:36
1
@jamesamajoe Not that the r-th summand is $$frac(2r colorred-1)^22^r$$ Y
– callculus
Jul 27 at 18:54
 |Â
show 6 more comments
everything is getting messed up as i m applying infinity while doing summation what to do where am i wrong
– jame samajoe
Jul 27 at 18:12
You... forgot the factor $1/2^r$ in your 3 resulting summations.
– Clement C.
Jul 27 at 18:32
@ClementC. can u explain what u want to say
– jame samajoe
Jul 27 at 18:35
You computed $$sum_r=1^n 4r^2 + sum_r=1^n 1+ sum_r=1^n 4r$$ instead of $$sum_r=1^n frac4r^22^r + sum_r=1^n frac12^r+ sum_r=1^n frac4r2^r$$
– Clement C.
Jul 27 at 18:36
1
@jamesamajoe Not that the r-th summand is $$frac(2r colorred-1)^22^r$$ Y
– callculus
Jul 27 at 18:54
everything is getting messed up as i m applying infinity while doing summation what to do where am i wrong
– jame samajoe
Jul 27 at 18:12
everything is getting messed up as i m applying infinity while doing summation what to do where am i wrong
– jame samajoe
Jul 27 at 18:12
You... forgot the factor $1/2^r$ in your 3 resulting summations.
– Clement C.
Jul 27 at 18:32
You... forgot the factor $1/2^r$ in your 3 resulting summations.
– Clement C.
Jul 27 at 18:32
@ClementC. can u explain what u want to say
– jame samajoe
Jul 27 at 18:35
@ClementC. can u explain what u want to say
– jame samajoe
Jul 27 at 18:35
You computed $$sum_r=1^n 4r^2 + sum_r=1^n 1+ sum_r=1^n 4r$$ instead of $$sum_r=1^n frac4r^22^r + sum_r=1^n frac12^r+ sum_r=1^n frac4r2^r$$
– Clement C.
Jul 27 at 18:36
You computed $$sum_r=1^n 4r^2 + sum_r=1^n 1+ sum_r=1^n 4r$$ instead of $$sum_r=1^n frac4r^22^r + sum_r=1^n frac12^r+ sum_r=1^n frac4r2^r$$
– Clement C.
Jul 27 at 18:36
1
1
@jamesamajoe Not that the r-th summand is $$frac(2r colorred-1)^22^r$$ Y
– callculus
Jul 27 at 18:54
@jamesamajoe Not that the r-th summand is $$frac(2r colorred-1)^22^r$$ Y
– callculus
Jul 27 at 18:54
 |Â
show 6 more comments
4 Answers
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up vote
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A simple answer to the question
Note that
beginmultline*
Sequivsum_ngeq1frac(2n-1)^22^n=sum_ngeq1frac4n^2-4n+12^n=4sum_ngeq1fracn^22^n-4sum_ngeq1fracn2^n+sum_ngeq1frac12^nequiv4I_2-4I_1+I_0.
endmultline*
since each of the series on the right hand side are convergent.
First, note that $I_0$ is a geometric series with $I_0=1$ (you can find an exposition to geometric series on Wikipedia).
As for $I_1$,
beginalign*
I_1 & =frac12^1+frac22^2+frac32^3+cdots\
frac12I_1 & =frac02^1+frac12^2+frac22^3+cdots\
frac12I_1=I_1-frac12I_1 & =frac12^1+frac12^2+frac12^3+cdots
endalign*
and hence $frac12I_1$ is once again a geometric series with $frac12I_1=1$ so that $I_1=2$.
As for $I_2$, note that
beginalign*
I_2 & =frac12^1+frac42^2+frac92^3+cdots\
frac12I_2 & =frac02^1+frac12^2+frac42^3+cdots\
frac12I_2=I_2-frac12I_2 & =frac12^1+frac32^2+frac52^3+cdots
endalign*
In other words,
$$
frac12I_2=sum_ngeq1frac2n-12^n=2I_1-I_0=2cdot2-1=3.
$$
Putting this all together,
$$
S=4cdot6-4cdot2+1=24-8+1=17.
$$
Generalizing the above
We can generalize the approach above. Fix a constant $cinmathbbC$ with $|c|>1$. For each nonnegative integer $m$, let
$$
boxedI_m=sum_ngeq1n^mc^-n
$$
Note that $I_0=(c-1)^-1$. If $m>0$, then
beginmultline*
left(c-1right)c^-1I_m=I_m-c^-1I_m=sum_ngeq1n^mc^-n-sum_ngeq1n^mc^-n-1\
=sum_ngeq1n^mc^-n-sum_ngeq1left(n-1right)^mc^-n=sum_ngeq1left(n^m-left(n-1right)^mright)c^-n\
=sum_ngeq1left(n^m-sum_k=0^mbinommkn^kleft(-1right)^m-kright)c^-n
=sum_ngeq1left(sum_k=0^m-1binommkn^kleft(-1right)^m-1-kright)c^-n\
=sum_k=0^m-1binommkleft(-1right)^m-1-kI_k.
endmultline*
This yields the recurrence
$$
boxedI_m=left(c-1right)^-1csum_k=0^m-1binommkleft(-1right)^m-1-kI_kqquadtextfor mgeq 1
$$
Relationship to polylogarithm
Note that $I_m$ is related to the polylogarithm:
$$
operatornameLi_-m(c^-1)=sum_ngeq1fracleft(c^-1right)^nn^-m=sum_ngeq1fracc^-nn^-m=sum_ngeq1n^mc^-n=I_m.
$$
Expressing our identity in terms of the polylogarithm,
$$
boxed<1
$$
There are various other expressions for $operatornameLi_-m(z)$
where $m$ is positive and $z$ is complex.
answered Jul 27 at 18:44
parsiad
15.9k32253
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Hint:
For $a<1$,
$$S_0(k):=sum_k=1^infty a^k=frac a1-a.$$
Then
$$S_1(k):=sum_k=1^infty ka^k=sum_k=1^infty a^k+asum_k=1^infty(k-1)a^k-1,$$
$$S_1(k)=S_0(k)+aS_1(k)$$ and
$$S_1(k)=frac a(1-a)^2.$$
Next,
$$S_2(k):=sum_k=1^infty k(k+1)a^k=sum_k=1^infty 2ka^k+asum_k=1^infty(k-1)ka^k-1,$$
$$S_2(k)=2S_1(k)+aS_2(k)$$ and
$$S_2(k)=frac2a(1-a)^3.$$
You can continue with $k(k+1)(k+2)a^ktodfrac3!a(1-a)^4$ and so on.
Now, any polynomial in $k$, such as $(2k+1)^2$ can be expressed as a linear combination of $1,k,k(k+1),cdots$ and the summation follows.
$$(2k-1)^2=4k(k+1)-8k+1,a=frac12to4fracfrac22frac12^3-8fracfrac12frac12^2+fracfrac12frac12=17.$$
answered Jul 27 at 21:28
Yves Daoust
110k665203
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The general term is $frac(2k-1)^22^k$. Using that, we get
$$
beginalign
sum_k=1^inftyfrac(2k-1)^22^k
&=sum_k=1^inftyfrac4k(k-1)+12^ktag1\
&=sum_k=1^infty8binomkk-2left(frac12right)^k+sum_k=1^inftybinomk-1k-1left(frac12right)^ktag2\
&=sum_k=1^infty8binom-3k-2left(-frac12right)^k-sum_k=1^inftybinom-1k-1left(-frac12right)^ktag3\
&=2left(1-frac12right)^-3+frac12left(1-frac12right)^-1tag4\[12pt]
&=17tag5
endalign
$$
Explanation:
$(1)$: $(2k-1)^2=4k(k-1)+1$
$(2)$: $4k(k-1)=8binomk2=8binomkk-2$ for $kge2$ and $binomk-1k-1=1$ for $kge1$
$(3)$: convert to negative binomial coefficients
$(4)$: Generalized Binomial Theorem
$(5)$: evaluate
answered Jul 27 at 21:21
robjohn♦
258k25297612
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Consider$$S_r=sum_n=1^r(2n-1)^2x^n=sum_n=1^r(4n^2-4n+1)x^n$$ Now, rewrite
$$n^2=n(n-1)+n$$ which makes
$$S_r=sum_n=1^r(4n(n-1)+1)x^n=4x^2sum_n=1^r n(n-1)x^n-2+sum_n=1^r x^n$$ that is to say $$S_r=4x^2 left(sum_n=1^r x^n right)''+sum_n=1^r x^n$$
Recall that $$sum_n=1^r x^n=fracx left(1-x^rright)1-x$$ Compute the second derivative and simplify; when finished, make $x=frac 12$ and take the limit for $r to infty$.
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
A simple answer to the question
Note that
beginmultline*
Sequivsum_ngeq1frac(2n-1)^22^n=sum_ngeq1frac4n^2-4n+12^n=4sum_ngeq1fracn^22^n-4sum_ngeq1fracn2^n+sum_ngeq1frac12^nequiv4I_2-4I_1+I_0.
endmultline*
since each of the series on the right hand side are convergent.
First, note that $I_0$ is a geometric series with $I_0=1$ (you can find an exposition to geometric series on Wikipedia).
As for $I_1$,
beginalign*
I_1 & =frac12^1+frac22^2+frac32^3+cdots\
frac12I_1 & =frac02^1+frac12^2+frac22^3+cdots\
frac12I_1=I_1-frac12I_1 & =frac12^1+frac12^2+frac12^3+cdots
endalign*
and hence $frac12I_1$ is once again a geometric series with $frac12I_1=1$ so that $I_1=2$.
As for $I_2$, note that
beginalign*
I_2 & =frac12^1+frac42^2+frac92^3+cdots\
frac12I_2 & =frac02^1+frac12^2+frac42^3+cdots\
frac12I_2=I_2-frac12I_2 & =frac12^1+frac32^2+frac52^3+cdots
endalign*
In other words,
$$
frac12I_2=sum_ngeq1frac2n-12^n=2I_1-I_0=2cdot2-1=3.
$$
Putting this all together,
$$
S=4cdot6-4cdot2+1=24-8+1=17.
$$
Generalizing the above
We can generalize the approach above. Fix a constant $cinmathbbC$ with $|c|>1$. For each nonnegative integer $m$, let
$$
boxedI_m=sum_ngeq1n^mc^-n
$$
Note that $I_0=(c-1)^-1$. If $m>0$, then
beginmultline*
left(c-1right)c^-1I_m=I_m-c^-1I_m=sum_ngeq1n^mc^-n-sum_ngeq1n^mc^-n-1\
=sum_ngeq1n^mc^-n-sum_ngeq1left(n-1right)^mc^-n=sum_ngeq1left(n^m-left(n-1right)^mright)c^-n\
=sum_ngeq1left(n^m-sum_k=0^mbinommkn^kleft(-1right)^m-kright)c^-n
=sum_ngeq1left(sum_k=0^m-1binommkn^kleft(-1right)^m-1-kright)c^-n\
=sum_k=0^m-1binommkleft(-1right)^m-1-kI_k.
endmultline*
This yields the recurrence
$$
boxedI_m=left(c-1right)^-1csum_k=0^m-1binommkleft(-1right)^m-1-kI_kqquadtextfor mgeq 1
$$
Relationship to polylogarithm
Note that $I_m$ is related to the polylogarithm:
$$
operatornameLi_-m(c^-1)=sum_ngeq1fracleft(c^-1right)^nn^-m=sum_ngeq1fracc^-nn^-m=sum_ngeq1n^mc^-n=I_m.
$$
Expressing our identity in terms of the polylogarithm,
$$
boxed<1
$$
There are various other expressions for $operatornameLi_-m(z)$
where $m$ is positive and $z$ is complex.
answered Jul 27 at 18:44
parsiad
15.9k32253
add a comment |Â
up vote
5
down vote
accepted
A simple answer to the question
Note that
beginmultline*
Sequivsum_ngeq1frac(2n-1)^22^n=sum_ngeq1frac4n^2-4n+12^n=4sum_ngeq1fracn^22^n-4sum_ngeq1fracn2^n+sum_ngeq1frac12^nequiv4I_2-4I_1+I_0.
endmultline*
since each of the series on the right hand side are convergent.
First, note that $I_0$ is a geometric series with $I_0=1$ (you can find an exposition to geometric series on Wikipedia).
As for $I_1$,
beginalign*
I_1 & =frac12^1+frac22^2+frac32^3+cdots\
frac12I_1 & =frac02^1+frac12^2+frac22^3+cdots\
frac12I_1=I_1-frac12I_1 & =frac12^1+frac12^2+frac12^3+cdots
endalign*
and hence $frac12I_1$ is once again a geometric series with $frac12I_1=1$ so that $I_1=2$.
As for $I_2$, note that
beginalign*
I_2 & =frac12^1+frac42^2+frac92^3+cdots\
frac12I_2 & =frac02^1+frac12^2+frac42^3+cdots\
frac12I_2=I_2-frac12I_2 & =frac12^1+frac32^2+frac52^3+cdots
endalign*
In other words,
$$
frac12I_2=sum_ngeq1frac2n-12^n=2I_1-I_0=2cdot2-1=3.
$$
Putting this all together,
$$
S=4cdot6-4cdot2+1=24-8+1=17.
$$
Generalizing the above
We can generalize the approach above. Fix a constant $cinmathbbC$ with $|c|>1$. For each nonnegative integer $m$, let
$$
boxedI_m=sum_ngeq1n^mc^-n
$$
Note that $I_0=(c-1)^-1$. If $m>0$, then
beginmultline*
left(c-1right)c^-1I_m=I_m-c^-1I_m=sum_ngeq1n^mc^-n-sum_ngeq1n^mc^-n-1\
=sum_ngeq1n^mc^-n-sum_ngeq1left(n-1right)^mc^-n=sum_ngeq1left(n^m-left(n-1right)^mright)c^-n\
=sum_ngeq1left(n^m-sum_k=0^mbinommkn^kleft(-1right)^m-kright)c^-n
=sum_ngeq1left(sum_k=0^m-1binommkn^kleft(-1right)^m-1-kright)c^-n\
=sum_k=0^m-1binommkleft(-1right)^m-1-kI_k.
endmultline*
This yields the recurrence
$$
boxedI_m=left(c-1right)^-1csum_k=0^m-1binommkleft(-1right)^m-1-kI_kqquadtextfor mgeq 1
$$
Relationship to polylogarithm
Note that $I_m$ is related to the polylogarithm:
$$
operatornameLi_-m(c^-1)=sum_ngeq1fracleft(c^-1right)^nn^-m=sum_ngeq1fracc^-nn^-m=sum_ngeq1n^mc^-n=I_m.
$$
Expressing our identity in terms of the polylogarithm,
$$
boxed<1
$$
There are various other expressions for $operatornameLi_-m(z)$
where $m$ is positive and $z$ is complex.
answered Jul 27 at 18:44
parsiad
15.9k32253
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
A simple answer to the question
Note that
beginmultline*
Sequivsum_ngeq1frac(2n-1)^22^n=sum_ngeq1frac4n^2-4n+12^n=4sum_ngeq1fracn^22^n-4sum_ngeq1fracn2^n+sum_ngeq1frac12^nequiv4I_2-4I_1+I_0.
endmultline*
since each of the series on the right hand side are convergent.
First, note that $I_0$ is a geometric series with $I_0=1$ (you can find an exposition to geometric series on Wikipedia).
As for $I_1$,
beginalign*
I_1 & =frac12^1+frac22^2+frac32^3+cdots\
frac12I_1 & =frac02^1+frac12^2+frac22^3+cdots\
frac12I_1=I_1-frac12I_1 & =frac12^1+frac12^2+frac12^3+cdots
endalign*
and hence $frac12I_1$ is once again a geometric series with $frac12I_1=1$ so that $I_1=2$.
As for $I_2$, note that
beginalign*
I_2 & =frac12^1+frac42^2+frac92^3+cdots\
frac12I_2 & =frac02^1+frac12^2+frac42^3+cdots\
frac12I_2=I_2-frac12I_2 & =frac12^1+frac32^2+frac52^3+cdots
endalign*
In other words,
$$
frac12I_2=sum_ngeq1frac2n-12^n=2I_1-I_0=2cdot2-1=3.
$$
Putting this all together,
$$
S=4cdot6-4cdot2+1=24-8+1=17.
$$
Generalizing the above
We can generalize the approach above. Fix a constant $cinmathbbC$ with $|c|>1$. For each nonnegative integer $m$, let
$$
boxedI_m=sum_ngeq1n^mc^-n
$$
Note that $I_0=(c-1)^-1$. If $m>0$, then
beginmultline*
left(c-1right)c^-1I_m=I_m-c^-1I_m=sum_ngeq1n^mc^-n-sum_ngeq1n^mc^-n-1\
=sum_ngeq1n^mc^-n-sum_ngeq1left(n-1right)^mc^-n=sum_ngeq1left(n^m-left(n-1right)^mright)c^-n\
=sum_ngeq1left(n^m-sum_k=0^mbinommkn^kleft(-1right)^m-kright)c^-n
=sum_ngeq1left(sum_k=0^m-1binommkn^kleft(-1right)^m-1-kright)c^-n\
=sum_k=0^m-1binommkleft(-1right)^m-1-kI_k.
endmultline*
This yields the recurrence
$$
boxedI_m=left(c-1right)^-1csum_k=0^m-1binommkleft(-1right)^m-1-kI_kqquadtextfor mgeq 1
$$
Relationship to polylogarithm
Note that $I_m$ is related to the polylogarithm:
$$
operatornameLi_-m(c^-1)=sum_ngeq1fracleft(c^-1right)^nn^-m=sum_ngeq1fracc^-nn^-m=sum_ngeq1n^mc^-n=I_m.
$$
Expressing our identity in terms of the polylogarithm,
$$
boxed<1
$$
There are various other expressions for $operatornameLi_-m(z)$
where $m$ is positive and $z$ is complex.
answered Jul 27 at 18:44
parsiad
15.9k32253
A simple answer to the question
Note that
beginmultline*
Sequivsum_ngeq1frac(2n-1)^22^n=sum_ngeq1frac4n^2-4n+12^n=4sum_ngeq1fracn^22^n-4sum_ngeq1fracn2^n+sum_ngeq1frac12^nequiv4I_2-4I_1+I_0.
endmultline*
since each of the series on the right hand side are convergent.
First, note that $I_0$ is a geometric series with $I_0=1$ (you can find an exposition to geometric series on Wikipedia).
As for $I_1$,
beginalign*
I_1 & =frac12^1+frac22^2+frac32^3+cdots\
frac12I_1 & =frac02^1+frac12^2+frac22^3+cdots\
frac12I_1=I_1-frac12I_1 & =frac12^1+frac12^2+frac12^3+cdots
endalign*
and hence $frac12I_1$ is once again a geometric series with $frac12I_1=1$ so that $I_1=2$.
As for $I_2$, note that
beginalign*
I_2 & =frac12^1+frac42^2+frac92^3+cdots\
frac12I_2 & =frac02^1+frac12^2+frac42^3+cdots\
frac12I_2=I_2-frac12I_2 & =frac12^1+frac32^2+frac52^3+cdots
endalign*
In other words,
$$
frac12I_2=sum_ngeq1frac2n-12^n=2I_1-I_0=2cdot2-1=3.
$$
Putting this all together,
$$
S=4cdot6-4cdot2+1=24-8+1=17.
$$
Generalizing the above
We can generalize the approach above. Fix a constant $cinmathbbC$ with $|c|>1$. For each nonnegative integer $m$, let
$$
boxedI_m=sum_ngeq1n^mc^-n
$$
Note that $I_0=(c-1)^-1$. If $m>0$, then
beginmultline*
left(c-1right)c^-1I_m=I_m-c^-1I_m=sum_ngeq1n^mc^-n-sum_ngeq1n^mc^-n-1\
=sum_ngeq1n^mc^-n-sum_ngeq1left(n-1right)^mc^-n=sum_ngeq1left(n^m-left(n-1right)^mright)c^-n\
=sum_ngeq1left(n^m-sum_k=0^mbinommkn^kleft(-1right)^m-kright)c^-n
=sum_ngeq1left(sum_k=0^m-1binommkn^kleft(-1right)^m-1-kright)c^-n\
=sum_k=0^m-1binommkleft(-1right)^m-1-kI_k.
endmultline*
This yields the recurrence
$$
boxedI_m=left(c-1right)^-1csum_k=0^m-1binommkleft(-1right)^m-1-kI_kqquadtextfor mgeq 1
$$
Relationship to polylogarithm
Note that $I_m$ is related to the polylogarithm:
$$
operatornameLi_-m(c^-1)=sum_ngeq1fracleft(c^-1right)^nn^-m=sum_ngeq1fracc^-nn^-m=sum_ngeq1n^mc^-n=I_m.
$$
Expressing our identity in terms of the polylogarithm,
$$
boxed<1
$$
There are various other expressions for $operatornameLi_-m(z)$
where $m$ is positive and $z$ is complex.
answered Jul 27 at 18:44
parsiad
15.9k32253
edited Jul 27 at 22:14
answered Jul 27 at 18:44
parsiad
15.9k32253
answered Jul 27 at 18:44
parsiad
15.9k32253
answered Jul 27 at 18:44
parsiad
15.9k32253
15.9k32253
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2
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Hint:
For $a<1$,
$$S_0(k):=sum_k=1^infty a^k=frac a1-a.$$
Then
$$S_1(k):=sum_k=1^infty ka^k=sum_k=1^infty a^k+asum_k=1^infty(k-1)a^k-1,$$
$$S_1(k)=S_0(k)+aS_1(k)$$ and
$$S_1(k)=frac a(1-a)^2.$$
Next,
$$S_2(k):=sum_k=1^infty k(k+1)a^k=sum_k=1^infty 2ka^k+asum_k=1^infty(k-1)ka^k-1,$$
$$S_2(k)=2S_1(k)+aS_2(k)$$ and
$$S_2(k)=frac2a(1-a)^3.$$
You can continue with $k(k+1)(k+2)a^ktodfrac3!a(1-a)^4$ and so on.
Now, any polynomial in $k$, such as $(2k+1)^2$ can be expressed as a linear combination of $1,k,k(k+1),cdots$ and the summation follows.
$$(2k-1)^2=4k(k+1)-8k+1,a=frac12to4fracfrac22frac12^3-8fracfrac12frac12^2+fracfrac12frac12=17.$$
answered Jul 27 at 21:28
Yves Daoust
110k665203
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up vote
2
down vote
Hint:
For $a<1$,
$$S_0(k):=sum_k=1^infty a^k=frac a1-a.$$
Then
$$S_1(k):=sum_k=1^infty ka^k=sum_k=1^infty a^k+asum_k=1^infty(k-1)a^k-1,$$
$$S_1(k)=S_0(k)+aS_1(k)$$ and
$$S_1(k)=frac a(1-a)^2.$$
Next,
$$S_2(k):=sum_k=1^infty k(k+1)a^k=sum_k=1^infty 2ka^k+asum_k=1^infty(k-1)ka^k-1,$$
$$S_2(k)=2S_1(k)+aS_2(k)$$ and
$$S_2(k)=frac2a(1-a)^3.$$
You can continue with $k(k+1)(k+2)a^ktodfrac3!a(1-a)^4$ and so on.
Now, any polynomial in $k$, such as $(2k+1)^2$ can be expressed as a linear combination of $1,k,k(k+1),cdots$ and the summation follows.
$$(2k-1)^2=4k(k+1)-8k+1,a=frac12to4fracfrac22frac12^3-8fracfrac12frac12^2+fracfrac12frac12=17.$$
answered Jul 27 at 21:28
Yves Daoust
110k665203
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
For $a<1$,
$$S_0(k):=sum_k=1^infty a^k=frac a1-a.$$
Then
$$S_1(k):=sum_k=1^infty ka^k=sum_k=1^infty a^k+asum_k=1^infty(k-1)a^k-1,$$
$$S_1(k)=S_0(k)+aS_1(k)$$ and
$$S_1(k)=frac a(1-a)^2.$$
Next,
$$S_2(k):=sum_k=1^infty k(k+1)a^k=sum_k=1^infty 2ka^k+asum_k=1^infty(k-1)ka^k-1,$$
$$S_2(k)=2S_1(k)+aS_2(k)$$ and
$$S_2(k)=frac2a(1-a)^3.$$
You can continue with $k(k+1)(k+2)a^ktodfrac3!a(1-a)^4$ and so on.
Now, any polynomial in $k$, such as $(2k+1)^2$ can be expressed as a linear combination of $1,k,k(k+1),cdots$ and the summation follows.
$$(2k-1)^2=4k(k+1)-8k+1,a=frac12to4fracfrac22frac12^3-8fracfrac12frac12^2+fracfrac12frac12=17.$$
answered Jul 27 at 21:28
Yves Daoust
110k665203
Hint:
For $a<1$,
$$S_0(k):=sum_k=1^infty a^k=frac a1-a.$$
Then
$$S_1(k):=sum_k=1^infty ka^k=sum_k=1^infty a^k+asum_k=1^infty(k-1)a^k-1,$$
$$S_1(k)=S_0(k)+aS_1(k)$$ and
$$S_1(k)=frac a(1-a)^2.$$
Next,
$$S_2(k):=sum_k=1^infty k(k+1)a^k=sum_k=1^infty 2ka^k+asum_k=1^infty(k-1)ka^k-1,$$
$$S_2(k)=2S_1(k)+aS_2(k)$$ and
$$S_2(k)=frac2a(1-a)^3.$$
You can continue with $k(k+1)(k+2)a^ktodfrac3!a(1-a)^4$ and so on.
Now, any polynomial in $k$, such as $(2k+1)^2$ can be expressed as a linear combination of $1,k,k(k+1),cdots$ and the summation follows.
$$(2k-1)^2=4k(k+1)-8k+1,a=frac12to4fracfrac22frac12^3-8fracfrac12frac12^2+fracfrac12frac12=17.$$
answered Jul 27 at 21:28
Yves Daoust
110k665203
edited Jul 27 at 21:40
answered Jul 27 at 21:28
Yves Daoust
110k665203
answered Jul 27 at 21:28
Yves Daoust
110k665203
answered Jul 27 at 21:28
Yves Daoust
110k665203
110k665203
add a comment |Â
add a comment |Â
up vote
2
down vote
The general term is $frac(2k-1)^22^k$. Using that, we get
$$
beginalign
sum_k=1^inftyfrac(2k-1)^22^k
&=sum_k=1^inftyfrac4k(k-1)+12^ktag1\
&=sum_k=1^infty8binomkk-2left(frac12right)^k+sum_k=1^inftybinomk-1k-1left(frac12right)^ktag2\
&=sum_k=1^infty8binom-3k-2left(-frac12right)^k-sum_k=1^inftybinom-1k-1left(-frac12right)^ktag3\
&=2left(1-frac12right)^-3+frac12left(1-frac12right)^-1tag4\[12pt]
&=17tag5
endalign
$$
Explanation:
$(1)$: $(2k-1)^2=4k(k-1)+1$
$(2)$: $4k(k-1)=8binomk2=8binomkk-2$ for $kge2$ and $binomk-1k-1=1$ for $kge1$
$(3)$: convert to negative binomial coefficients
$(4)$: Generalized Binomial Theorem
$(5)$: evaluate
answered Jul 27 at 21:21
robjohn♦
258k25297612
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up vote
2
down vote
The general term is $frac(2k-1)^22^k$. Using that, we get
$$
beginalign
sum_k=1^inftyfrac(2k-1)^22^k
&=sum_k=1^inftyfrac4k(k-1)+12^ktag1\
&=sum_k=1^infty8binomkk-2left(frac12right)^k+sum_k=1^inftybinomk-1k-1left(frac12right)^ktag2\
&=sum_k=1^infty8binom-3k-2left(-frac12right)^k-sum_k=1^inftybinom-1k-1left(-frac12right)^ktag3\
&=2left(1-frac12right)^-3+frac12left(1-frac12right)^-1tag4\[12pt]
&=17tag5
endalign
$$
Explanation:
$(1)$: $(2k-1)^2=4k(k-1)+1$
$(2)$: $4k(k-1)=8binomk2=8binomkk-2$ for $kge2$ and $binomk-1k-1=1$ for $kge1$
$(3)$: convert to negative binomial coefficients
$(4)$: Generalized Binomial Theorem
$(5)$: evaluate
answered Jul 27 at 21:21
robjohn♦
258k25297612
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up vote
2
down vote
up vote
2
down vote
The general term is $frac(2k-1)^22^k$. Using that, we get
$$
beginalign
sum_k=1^inftyfrac(2k-1)^22^k
&=sum_k=1^inftyfrac4k(k-1)+12^ktag1\
&=sum_k=1^infty8binomkk-2left(frac12right)^k+sum_k=1^inftybinomk-1k-1left(frac12right)^ktag2\
&=sum_k=1^infty8binom-3k-2left(-frac12right)^k-sum_k=1^inftybinom-1k-1left(-frac12right)^ktag3\
&=2left(1-frac12right)^-3+frac12left(1-frac12right)^-1tag4\[12pt]
&=17tag5
endalign
$$
Explanation:
$(1)$: $(2k-1)^2=4k(k-1)+1$
$(2)$: $4k(k-1)=8binomk2=8binomkk-2$ for $kge2$ and $binomk-1k-1=1$ for $kge1$
$(3)$: convert to negative binomial coefficients
$(4)$: Generalized Binomial Theorem
$(5)$: evaluate
answered Jul 27 at 21:21
robjohn♦
258k25297612
The general term is $frac(2k-1)^22^k$. Using that, we get
$$
beginalign
sum_k=1^inftyfrac(2k-1)^22^k
&=sum_k=1^inftyfrac4k(k-1)+12^ktag1\
&=sum_k=1^infty8binomkk-2left(frac12right)^k+sum_k=1^inftybinomk-1k-1left(frac12right)^ktag2\
&=sum_k=1^infty8binom-3k-2left(-frac12right)^k-sum_k=1^inftybinom-1k-1left(-frac12right)^ktag3\
&=2left(1-frac12right)^-3+frac12left(1-frac12right)^-1tag4\[12pt]
&=17tag5
endalign
$$
Explanation:
$(1)$: $(2k-1)^2=4k(k-1)+1$
$(2)$: $4k(k-1)=8binomk2=8binomkk-2$ for $kge2$ and $binomk-1k-1=1$ for $kge1$
$(3)$: convert to negative binomial coefficients
$(4)$: Generalized Binomial Theorem
$(5)$: evaluate
answered Jul 27 at 21:21
robjohn♦
258k25297612
edited Jul 27 at 22:22
answered Jul 27 at 21:21
robjohn♦
258k25297612
answered Jul 27 at 21:21
robjohn♦
258k25297612
answered Jul 27 at 21:21
robjohn♦
258k25297612
258k25297612
add a comment |Â
add a comment |Â
up vote
1
down vote
Consider$$S_r=sum_n=1^r(2n-1)^2x^n=sum_n=1^r(4n^2-4n+1)x^n$$ Now, rewrite
$$n^2=n(n-1)+n$$ which makes
$$S_r=sum_n=1^r(4n(n-1)+1)x^n=4x^2sum_n=1^r n(n-1)x^n-2+sum_n=1^r x^n$$ that is to say $$S_r=4x^2 left(sum_n=1^r x^n right)''+sum_n=1^r x^n$$
Recall that $$sum_n=1^r x^n=fracx left(1-x^rright)1-x$$ Compute the second derivative and simplify; when finished, make $x=frac 12$ and take the limit for $r to infty$.
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
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up vote
1
down vote
Consider$$S_r=sum_n=1^r(2n-1)^2x^n=sum_n=1^r(4n^2-4n+1)x^n$$ Now, rewrite
$$n^2=n(n-1)+n$$ which makes
$$S_r=sum_n=1^r(4n(n-1)+1)x^n=4x^2sum_n=1^r n(n-1)x^n-2+sum_n=1^r x^n$$ that is to say $$S_r=4x^2 left(sum_n=1^r x^n right)''+sum_n=1^r x^n$$
Recall that $$sum_n=1^r x^n=fracx left(1-x^rright)1-x$$ Compute the second derivative and simplify; when finished, make $x=frac 12$ and take the limit for $r to infty$.
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Consider$$S_r=sum_n=1^r(2n-1)^2x^n=sum_n=1^r(4n^2-4n+1)x^n$$ Now, rewrite
$$n^2=n(n-1)+n$$ which makes
$$S_r=sum_n=1^r(4n(n-1)+1)x^n=4x^2sum_n=1^r n(n-1)x^n-2+sum_n=1^r x^n$$ that is to say $$S_r=4x^2 left(sum_n=1^r x^n right)''+sum_n=1^r x^n$$
Recall that $$sum_n=1^r x^n=fracx left(1-x^rright)1-x$$ Compute the second derivative and simplify; when finished, make $x=frac 12$ and take the limit for $r to infty$.
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
Consider$$S_r=sum_n=1^r(2n-1)^2x^n=sum_n=1^r(4n^2-4n+1)x^n$$ Now, rewrite
$$n^2=n(n-1)+n$$ which makes
$$S_r=sum_n=1^r(4n(n-1)+1)x^n=4x^2sum_n=1^r n(n-1)x^n-2+sum_n=1^r x^n$$ that is to say $$S_r=4x^2 left(sum_n=1^r x^n right)''+sum_n=1^r x^n$$
Recall that $$sum_n=1^r x^n=fracx left(1-x^rright)1-x$$ Compute the second derivative and simplify; when finished, make $x=frac 12$ and take the limit for $r to infty$.
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
answered Jul 28 at 6:04
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
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everything is getting messed up as i m applying infinity while doing summation what to do where am i wrong
– jame samajoe
Jul 27 at 18:12
You... forgot the factor $1/2^r$ in your 3 resulting summations.
– Clement C.
Jul 27 at 18:32
@ClementC. can u explain what u want to say
– jame samajoe
Jul 27 at 18:35
You computed $$sum_r=1^n 4r^2 + sum_r=1^n 1+ sum_r=1^n 4r$$ instead of $$sum_r=1^n frac4r^22^r + sum_r=1^n frac12^r+ sum_r=1^n frac4r2^r$$
– Clement C.
Jul 27 at 18:36
1
@jamesamajoe Not that the r-th summand is $$frac(2r colorred-1)^22^r$$ Y
– callculus
Jul 27 at 18:54