Mixing
$frac1n+1+frac1(n+1)^2…=frac1n$?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
On a problem book solution I was faced with the following step:
$$frac1n+1+frac1(n+1)^2...=frac1n$$
I identified $frac1n+1+frac1(n+1)^2...$ as a geometric series so the sum would be $frac11-r$ so that $frac11-frac1n+1=1+frac1n$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!
calculus sequences-and-series
edited Jul 31 at 22:08
amWhy
189k25219431
asked Jul 31 at 21:41
Pedro Gomes
1,3002618
add a comment |Â
up vote
1
down vote
favorite
On a problem book solution I was faced with the following step:
$$frac1n+1+frac1(n+1)^2...=frac1n$$
I identified $frac1n+1+frac1(n+1)^2...$ as a geometric series so the sum would be $frac11-r$ so that $frac11-frac1n+1=1+frac1n$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!
calculus sequences-and-series
edited Jul 31 at 22:08
amWhy
189k25219431
asked Jul 31 at 21:41
Pedro Gomes
1,3002618
6
It's not a whole geometric series. There is a first term of $1$ before the $dfrac1n+1$ term. Add that in, the $1$s on both sides cancel out to get the equation you want.
– user496634
Jul 31 at 21:43
4
It's a geometric series starting at $k=1$ not $k=0$
– mrnoqwerty
Jul 31 at 21:43
So the sum of $ sum_k=0^infty r^k = frac 11-r=1+frac 1n$ so $sum_k=1^infty r^k = ( sum_k=0^infty r^k ) - r^0 = ....????....$
– fleablood
Jul 31 at 22:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On a problem book solution I was faced with the following step:
$$frac1n+1+frac1(n+1)^2...=frac1n$$
I identified $frac1n+1+frac1(n+1)^2...$ as a geometric series so the sum would be $frac11-r$ so that $frac11-frac1n+1=1+frac1n$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!
calculus sequences-and-series
edited Jul 31 at 22:08
amWhy
189k25219431
asked Jul 31 at 21:41
Pedro Gomes
1,3002618
On a problem book solution I was faced with the following step:
$$frac1n+1+frac1(n+1)^2...=frac1n$$
I identified $frac1n+1+frac1(n+1)^2...$ as a geometric series so the sum would be $frac11-r$ so that $frac11-frac1n+1=1+frac1n$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!
calculus sequences-and-series
edited Jul 31 at 22:08
amWhy
189k25219431
asked Jul 31 at 21:41
Pedro Gomes
1,3002618
edited Jul 31 at 22:08
amWhy
189k25219431
edited Jul 31 at 22:08
amWhy
189k25219431
edited Jul 31 at 22:08
amWhy
189k25219431
189k25219431
asked Jul 31 at 21:41
Pedro Gomes
1,3002618
asked Jul 31 at 21:41
Pedro Gomes
1,3002618
asked Jul 31 at 21:41
Pedro Gomes
1,3002618
1,3002618
6
It's not a whole geometric series. There is a first term of $1$ before the $dfrac1n+1$ term. Add that in, the $1$s on both sides cancel out to get the equation you want.
– user496634
Jul 31 at 21:43
4
It's a geometric series starting at $k=1$ not $k=0$
– mrnoqwerty
Jul 31 at 21:43
So the sum of $ sum_k=0^infty r^k = frac 11-r=1+frac 1n$ so $sum_k=1^infty r^k = ( sum_k=0^infty r^k ) - r^0 = ....????....$
– fleablood
Jul 31 at 22:01
add a comment |Â
6
It's not a whole geometric series. There is a first term of $1$ before the $dfrac1n+1$ term. Add that in, the $1$s on both sides cancel out to get the equation you want.
– user496634
Jul 31 at 21:43
4
It's a geometric series starting at $k=1$ not $k=0$
– mrnoqwerty
Jul 31 at 21:43
So the sum of $ sum_k=0^infty r^k = frac 11-r=1+frac 1n$ so $sum_k=1^infty r^k = ( sum_k=0^infty r^k ) - r^0 = ....????....$
– fleablood
Jul 31 at 22:01
6
6
It's not a whole geometric series. There is a first term of $1$ before the $dfrac1n+1$ term. Add that in, the $1$s on both sides cancel out to get the equation you want.
– user496634
Jul 31 at 21:43
It's not a whole geometric series. There is a first term of $1$ before the $dfrac1n+1$ term. Add that in, the $1$s on both sides cancel out to get the equation you want.
– user496634
Jul 31 at 21:43
4
4
It's a geometric series starting at $k=1$ not $k=0$
– mrnoqwerty
Jul 31 at 21:43
It's a geometric series starting at $k=1$ not $k=0$
– mrnoqwerty
Jul 31 at 21:43
So the sum of $ sum_k=0^infty r^k = frac 11-r=1+frac 1n$ so $sum_k=1^infty r^k = ( sum_k=0^infty r^k ) - r^0 = ....????....$
– fleablood
Jul 31 at 22:01
So the sum of $ sum_k=0^infty r^k = frac 11-r=1+frac 1n$ so $sum_k=1^infty r^k = ( sum_k=0^infty r^k ) - r^0 = ....????....$
– fleablood
Jul 31 at 22:01
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
6
down vote
accepted
You are using the formula
$$
1+x+x^2+x^3+dotsb=frac11-xquad (|x|<1)quad (star)
$$
where $x=frac1n+1$. The problem is that your desired sum omits the initial term of $1$ in $(star)$. Hence subtract $1$ from your sum, $1+n^-1$, to get the right result.
answered Jul 31 at 21:50
Foobaz John
18k41245
add a comment |Â
up vote
3
down vote
The geometric sequence does not start with $1$.
The general form is $$ a + ar + ar^2 + ... = frac a1-r$$ for $|r|<1$
Therefor the sum
$$frac1n+1+frac1(n+1)^2...=frac1n$$
is correct.
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
If you know that $a+ar+ar^2+ar^3+cdots = dfraca1-r$ for $|r|lt 1$
then here you have $a= dfrac1n+1$ and $r= dfrac1n+1$
so the sum is $dfracfrac1n+11-frac1n+1=dfrac1n+1-1=dfrac1n$
answered Jul 31 at 21:52
Henry
92.8k469147
add a comment |Â
up vote
2
down vote
How I recommend approaching these problems:
Use the fact that if $|x| < 1$,
$$1 + x + x^2 + cdots = dfrac11-xtext. tag*$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $dfrac1n+1$ that
$$dfrac1n+1 + dfrac1(n+1)^2 + cdots = dfrac1n+1left(1+dfrac1n+1+dfrac1(n+1)^2+cdots right)text.tagA$$
The sum
$$1+dfrac1n+1+dfrac1(n+1)^2 + cdots = dfrac11-frac1n+1 = dfrac1(n+1-1)/(n+1) = dfracn+1ntagB$$
as long as $$left|dfrac1n+1 right| < 1 implies |n+1| > 1 implies n+1 > 1 text and -(n+1) < -1 implies n > 0text.$$
Thus, as long as $n > 0$, combine (A) and (B) to get
$$dfrac1n+1 cdot dfracn+1n = dfrac1n$$
as desired.
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
answered Jul 31 at 21:57
Clarinetist
10.3k32767
add a comment |Â
up vote
1
down vote
I identified 1n+1+1(n+1)2... as a geometric series so the sum would be 11−r so that 11−1n+1=1+1n. I do not understand what I am doing worng.
Take a look at your indexes.
A) You have $frac 11+n + (frac 1n+1)^2 + ..... = sum_k= 1^infty (frac 11+n)^k$
B) So you did: Let $r = frac 1n+1$ and $sum_k=0^infty( frac 11+n)^k =sum_k=0^inftyr^k=frac 11-r = 1 + frac 1n$.
Take a closer look do you see the difference between $sum_k= 1^infty (frac 11+n)^k$ in A) and $sum_k=0^infty (frac 11+n)^k$ in B)?
....
Now $sum_k= 1^infty frac 11+n^k = [sum_k=0^infty (frac 11+n)^k] - (frac 1n+1)^0 = ....???....$
answered Jul 31 at 22:09
fleablood
60.1k22575
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
You are using the formula
$$
1+x+x^2+x^3+dotsb=frac11-xquad (|x|<1)quad (star)
$$
where $x=frac1n+1$. The problem is that your desired sum omits the initial term of $1$ in $(star)$. Hence subtract $1$ from your sum, $1+n^-1$, to get the right result.
answered Jul 31 at 21:50
Foobaz John
18k41245
add a comment |Â
up vote
6
down vote
accepted
You are using the formula
$$
1+x+x^2+x^3+dotsb=frac11-xquad (|x|<1)quad (star)
$$
where $x=frac1n+1$. The problem is that your desired sum omits the initial term of $1$ in $(star)$. Hence subtract $1$ from your sum, $1+n^-1$, to get the right result.
answered Jul 31 at 21:50
Foobaz John
18k41245
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
You are using the formula
$$
1+x+x^2+x^3+dotsb=frac11-xquad (|x|<1)quad (star)
$$
where $x=frac1n+1$. The problem is that your desired sum omits the initial term of $1$ in $(star)$. Hence subtract $1$ from your sum, $1+n^-1$, to get the right result.
answered Jul 31 at 21:50
Foobaz John
18k41245
You are using the formula
$$
1+x+x^2+x^3+dotsb=frac11-xquad (|x|<1)quad (star)
$$
where $x=frac1n+1$. The problem is that your desired sum omits the initial term of $1$ in $(star)$. Hence subtract $1$ from your sum, $1+n^-1$, to get the right result.
answered Jul 31 at 21:50
Foobaz John
18k41245
answered Jul 31 at 21:50
Foobaz John
18k41245
answered Jul 31 at 21:50
Foobaz John
18k41245
answered Jul 31 at 21:50
Foobaz John
18k41245
18k41245
add a comment |Â
add a comment |Â
up vote
3
down vote
The geometric sequence does not start with $1$.
The general form is $$ a + ar + ar^2 + ... = frac a1-r$$ for $|r|<1$
Therefor the sum
$$frac1n+1+frac1(n+1)^2...=frac1n$$
is correct.
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
3
down vote
The geometric sequence does not start with $1$.
The general form is $$ a + ar + ar^2 + ... = frac a1-r$$ for $|r|<1$
Therefor the sum
$$frac1n+1+frac1(n+1)^2...=frac1n$$
is correct.
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The geometric sequence does not start with $1$.
The general form is $$ a + ar + ar^2 + ... = frac a1-r$$ for $|r|<1$
Therefor the sum
$$frac1n+1+frac1(n+1)^2...=frac1n$$
is correct.
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
The geometric sequence does not start with $1$.
The general form is $$ a + ar + ar^2 + ... = frac a1-r$$ for $|r|<1$
Therefor the sum
$$frac1n+1+frac1(n+1)^2...=frac1n$$
is correct.
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
answered Jul 31 at 21:50
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
up vote
2
down vote
If you know that $a+ar+ar^2+ar^3+cdots = dfraca1-r$ for $|r|lt 1$
then here you have $a= dfrac1n+1$ and $r= dfrac1n+1$
so the sum is $dfracfrac1n+11-frac1n+1=dfrac1n+1-1=dfrac1n$
answered Jul 31 at 21:52
Henry
92.8k469147
add a comment |Â
up vote
2
down vote
If you know that $a+ar+ar^2+ar^3+cdots = dfraca1-r$ for $|r|lt 1$
then here you have $a= dfrac1n+1$ and $r= dfrac1n+1$
so the sum is $dfracfrac1n+11-frac1n+1=dfrac1n+1-1=dfrac1n$
answered Jul 31 at 21:52
Henry
92.8k469147
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you know that $a+ar+ar^2+ar^3+cdots = dfraca1-r$ for $|r|lt 1$
then here you have $a= dfrac1n+1$ and $r= dfrac1n+1$
so the sum is $dfracfrac1n+11-frac1n+1=dfrac1n+1-1=dfrac1n$
answered Jul 31 at 21:52
Henry
92.8k469147
If you know that $a+ar+ar^2+ar^3+cdots = dfraca1-r$ for $|r|lt 1$
then here you have $a= dfrac1n+1$ and $r= dfrac1n+1$
so the sum is $dfracfrac1n+11-frac1n+1=dfrac1n+1-1=dfrac1n$
answered Jul 31 at 21:52
Henry
92.8k469147
answered Jul 31 at 21:52
Henry
92.8k469147
answered Jul 31 at 21:52
Henry
92.8k469147
answered Jul 31 at 21:52
Henry
92.8k469147
92.8k469147
add a comment |Â
add a comment |Â
up vote
2
down vote
How I recommend approaching these problems:
Use the fact that if $|x| < 1$,
$$1 + x + x^2 + cdots = dfrac11-xtext. tag*$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $dfrac1n+1$ that
$$dfrac1n+1 + dfrac1(n+1)^2 + cdots = dfrac1n+1left(1+dfrac1n+1+dfrac1(n+1)^2+cdots right)text.tagA$$
The sum
$$1+dfrac1n+1+dfrac1(n+1)^2 + cdots = dfrac11-frac1n+1 = dfrac1(n+1-1)/(n+1) = dfracn+1ntagB$$
as long as $$left|dfrac1n+1 right| < 1 implies |n+1| > 1 implies n+1 > 1 text and -(n+1) < -1 implies n > 0text.$$
Thus, as long as $n > 0$, combine (A) and (B) to get
$$dfrac1n+1 cdot dfracn+1n = dfrac1n$$
as desired.
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
answered Jul 31 at 21:57
Clarinetist
10.3k32767
add a comment |Â
up vote
2
down vote
How I recommend approaching these problems:
Use the fact that if $|x| < 1$,
$$1 + x + x^2 + cdots = dfrac11-xtext. tag*$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $dfrac1n+1$ that
$$dfrac1n+1 + dfrac1(n+1)^2 + cdots = dfrac1n+1left(1+dfrac1n+1+dfrac1(n+1)^2+cdots right)text.tagA$$
The sum
$$1+dfrac1n+1+dfrac1(n+1)^2 + cdots = dfrac11-frac1n+1 = dfrac1(n+1-1)/(n+1) = dfracn+1ntagB$$
as long as $$left|dfrac1n+1 right| < 1 implies |n+1| > 1 implies n+1 > 1 text and -(n+1) < -1 implies n > 0text.$$
Thus, as long as $n > 0$, combine (A) and (B) to get
$$dfrac1n+1 cdot dfracn+1n = dfrac1n$$
as desired.
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
answered Jul 31 at 21:57
Clarinetist
10.3k32767
add a comment |Â
up vote
2
down vote
up vote
2
down vote
How I recommend approaching these problems:
Use the fact that if $|x| < 1$,
$$1 + x + x^2 + cdots = dfrac11-xtext. tag*$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $dfrac1n+1$ that
$$dfrac1n+1 + dfrac1(n+1)^2 + cdots = dfrac1n+1left(1+dfrac1n+1+dfrac1(n+1)^2+cdots right)text.tagA$$
The sum
$$1+dfrac1n+1+dfrac1(n+1)^2 + cdots = dfrac11-frac1n+1 = dfrac1(n+1-1)/(n+1) = dfracn+1ntagB$$
as long as $$left|dfrac1n+1 right| < 1 implies |n+1| > 1 implies n+1 > 1 text and -(n+1) < -1 implies n > 0text.$$
Thus, as long as $n > 0$, combine (A) and (B) to get
$$dfrac1n+1 cdot dfracn+1n = dfrac1n$$
as desired.
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
answered Jul 31 at 21:57
Clarinetist
10.3k32767
How I recommend approaching these problems:
Use the fact that if $|x| < 1$,
$$1 + x + x^2 + cdots = dfrac11-xtext. tag*$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $dfrac1n+1$ that
$$dfrac1n+1 + dfrac1(n+1)^2 + cdots = dfrac1n+1left(1+dfrac1n+1+dfrac1(n+1)^2+cdots right)text.tagA$$
The sum
$$1+dfrac1n+1+dfrac1(n+1)^2 + cdots = dfrac11-frac1n+1 = dfrac1(n+1-1)/(n+1) = dfracn+1ntagB$$
as long as $$left|dfrac1n+1 right| < 1 implies |n+1| > 1 implies n+1 > 1 text and -(n+1) < -1 implies n > 0text.$$
Thus, as long as $n > 0$, combine (A) and (B) to get
$$dfrac1n+1 cdot dfracn+1n = dfrac1n$$
as desired.
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
answered Jul 31 at 21:57
Clarinetist
10.3k32767
answered Jul 31 at 21:57
Clarinetist
10.3k32767
answered Jul 31 at 21:57
Clarinetist
10.3k32767
answered Jul 31 at 21:57
Clarinetist
10.3k32767
10.3k32767
add a comment |Â
add a comment |Â
up vote
1
down vote
I identified 1n+1+1(n+1)2... as a geometric series so the sum would be 11−r so that 11−1n+1=1+1n. I do not understand what I am doing worng.
Take a look at your indexes.
A) You have $frac 11+n + (frac 1n+1)^2 + ..... = sum_k= 1^infty (frac 11+n)^k$
B) So you did: Let $r = frac 1n+1$ and $sum_k=0^infty( frac 11+n)^k =sum_k=0^inftyr^k=frac 11-r = 1 + frac 1n$.
Take a closer look do you see the difference between $sum_k= 1^infty (frac 11+n)^k$ in A) and $sum_k=0^infty (frac 11+n)^k$ in B)?
....
Now $sum_k= 1^infty frac 11+n^k = [sum_k=0^infty (frac 11+n)^k] - (frac 1n+1)^0 = ....???....$
answered Jul 31 at 22:09
fleablood
60.1k22575
add a comment |Â
up vote
1
down vote
I identified 1n+1+1(n+1)2... as a geometric series so the sum would be 11−r so that 11−1n+1=1+1n. I do not understand what I am doing worng.
Take a look at your indexes.
A) You have $frac 11+n + (frac 1n+1)^2 + ..... = sum_k= 1^infty (frac 11+n)^k$
B) So you did: Let $r = frac 1n+1$ and $sum_k=0^infty( frac 11+n)^k =sum_k=0^inftyr^k=frac 11-r = 1 + frac 1n$.
Take a closer look do you see the difference between $sum_k= 1^infty (frac 11+n)^k$ in A) and $sum_k=0^infty (frac 11+n)^k$ in B)?
....
Now $sum_k= 1^infty frac 11+n^k = [sum_k=0^infty (frac 11+n)^k] - (frac 1n+1)^0 = ....???....$
answered Jul 31 at 22:09
fleablood
60.1k22575
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I identified 1n+1+1(n+1)2... as a geometric series so the sum would be 11−r so that 11−1n+1=1+1n. I do not understand what I am doing worng.
Take a look at your indexes.
A) You have $frac 11+n + (frac 1n+1)^2 + ..... = sum_k= 1^infty (frac 11+n)^k$
B) So you did: Let $r = frac 1n+1$ and $sum_k=0^infty( frac 11+n)^k =sum_k=0^inftyr^k=frac 11-r = 1 + frac 1n$.
Take a closer look do you see the difference between $sum_k= 1^infty (frac 11+n)^k$ in A) and $sum_k=0^infty (frac 11+n)^k$ in B)?
....
Now $sum_k= 1^infty frac 11+n^k = [sum_k=0^infty (frac 11+n)^k] - (frac 1n+1)^0 = ....???....$
answered Jul 31 at 22:09
fleablood
60.1k22575
I identified 1n+1+1(n+1)2... as a geometric series so the sum would be 11−r so that 11−1n+1=1+1n. I do not understand what I am doing worng.
Take a look at your indexes.
A) You have $frac 11+n + (frac 1n+1)^2 + ..... = sum_k= 1^infty (frac 11+n)^k$
B) So you did: Let $r = frac 1n+1$ and $sum_k=0^infty( frac 11+n)^k =sum_k=0^inftyr^k=frac 11-r = 1 + frac 1n$.
Take a closer look do you see the difference between $sum_k= 1^infty (frac 11+n)^k$ in A) and $sum_k=0^infty (frac 11+n)^k$ in B)?
....
Now $sum_k= 1^infty frac 11+n^k = [sum_k=0^infty (frac 11+n)^k] - (frac 1n+1)^0 = ....???....$
answered Jul 31 at 22:09
fleablood
60.1k22575
answered Jul 31 at 22:09
fleablood
60.1k22575
answered Jul 31 at 22:09
fleablood
60.1k22575
answered Jul 31 at 22:09
fleablood
60.1k22575
60.1k22575
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868522%2ffrac1n1-frac1n12-frac1n%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
6
It's not a whole geometric series. There is a first term of $1$ before the $dfrac1n+1$ term. Add that in, the $1$s on both sides cancel out to get the equation you want.
– user496634
Jul 31 at 21:43
4
It's a geometric series starting at $k=1$ not $k=0$
– mrnoqwerty
Jul 31 at 21:43
So the sum of $ sum_k=0^infty r^k = frac 11-r=1+frac 1n$ so $sum_k=1^infty r^k = ( sum_k=0^infty r^k ) - r^0 = ....????....$
– fleablood
Jul 31 at 22:01