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Calculating sum of infinite series
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I am really confused by sum of infinite series and was hoping to get some guidance. If I have a series $$sum_i=1^infty fracii+1$$
If I am asked to calculate the sum of the series, do I just take the limit?
If I take the limit it is 1, which I think means that series diverges so there is no sum.
Or is it the case that the sum is 1? Does a series only diverge if the limit is infinity or does it diverge if the limit doesn't equal 0?
sequences-and-series
asked 1 hour ago
Sam
1136
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up vote
3
down vote
favorite
I am really confused by sum of infinite series and was hoping to get some guidance. If I have a series $$sum_i=1^infty fracii+1$$
If I am asked to calculate the sum of the series, do I just take the limit?
If I take the limit it is 1, which I think means that series diverges so there is no sum.
Or is it the case that the sum is 1? Does a series only diverge if the limit is infinity or does it diverge if the limit doesn't equal 0?
sequences-and-series
asked 1 hour ago
Sam
1136
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am really confused by sum of infinite series and was hoping to get some guidance. If I have a series $$sum_i=1^infty fracii+1$$
If I am asked to calculate the sum of the series, do I just take the limit?
If I take the limit it is 1, which I think means that series diverges so there is no sum.
Or is it the case that the sum is 1? Does a series only diverge if the limit is infinity or does it diverge if the limit doesn't equal 0?
sequences-and-series
asked 1 hour ago
Sam
1136
I am really confused by sum of infinite series and was hoping to get some guidance. If I have a series $$sum_i=1^infty fracii+1$$
If I am asked to calculate the sum of the series, do I just take the limit?
If I take the limit it is 1, which I think means that series diverges so there is no sum.
Or is it the case that the sum is 1? Does a series only diverge if the limit is infinity or does it diverge if the limit doesn't equal 0?
sequences-and-series
sequences-and-series
asked 1 hour ago
Sam
1136
asked 1 hour ago
Sam
1136
edited 59 mins ago
asked 1 hour ago
Sam
1136
asked 1 hour ago
Sam
1136
asked 1 hour ago
Sam
1136
1136
add a comment |Â
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
2
down vote
accepted
Ok, let's start from the very beginning, to try to clarify your doubts.
The writing $sum_i=1^inftyfracii+1$ doesn't mean that you are actually performing a sum of an infinite number of elements. It's a notation which stands for
$$lim_nrightarrowinftysum_i=1^nfracii+1.$$
This means that you are creating a sequence $sum_i=1^nfracii+1_n=1,2,...$
called sequence of partial sums and you are studying its limit for $n$ approaching $infty$. The series $sum_i=1^inftyfracii+1$ is said convergent if the sequence of partial sums converges.
Now when you say 'if I take the limit it is $1$' you are not referring to the limit of the parial sums. You are talking about $lim_irightarrowinftyfracii+1$. But why?
Well, this is because you have a result which tells you that if you have a series of the form
$sum_i=1^inftya_i$ which converges, then $lim_irightarrowinftya_i=0$.
In your case $a_i=fracii+1$ and you have $lim_irightarrowinftyfracii+1=1$. This means that your series is not convergent. As explained in the other answers the sequence of parial sums is going to $infty$ and the point is exactly that $fracii+1$ is approaching $1$.
Now the second part. Having $lim_irightarrowinftya_i=0$ is not enough to guaratee that the series $sum_i=1^infty a_i$ is convergent. And the standard example is the one you considered in one of the comments, namely $sum_i=1^inftyfrac1i$. For this series you have $lim_irightarrowinftyfrac1i=0$ but nonetheless the series is divergent.
answered 39 mins ago
Uskebasi
9021417
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up vote
2
down vote
A series is a sequence of partial sums. In your case,
$$frac12,frac12+frac23,frac12+frac23+frac34,frac12+frac23+frac34+frac45,cdots$$
or
$$frac12,frac76,frac2312,frac16360,cdots$$
Does this sequence converge ?
Hint:
$$frac ii+1gefrac12implies S_n>frac n2$$
answered 55 mins ago
Yves Daoust
115k666209
add a comment |Â
up vote
1
down vote
It means that the sum diverges. You are summing numbers that are larger than $frac12$, meaning that the sum of the first $n$ terms will be at least $fracn2$. That is, the partial sums grow without bound. As the sum of an infinite series is defined to be the limit of the partial sums, the series diverges.
answered 1 hour ago
Theo Bendit
13.7k12045
If this series did not diverge, is there a way to find the sum of that series? For example if the series of $sum_i=0^infty frac1n$ then the sum is 0, but is there a way to have a sum other than 0?
– Sam
1 hour ago
I'm not sure what you mean. The series does diverge, and any method that begins by assuming the series converges will arrive at totally erroneous conclusions.
– Theo Bendit
56 mins ago
add a comment |Â
up vote
1
down vote
You seem to be confusing the sequence $$a_i = frac ii+1$$
And the sequence of partial sums
$$s_n=sum_i=1^na_i$$
Now $$sum_i=1^infty fracii+1$$
converges iff $s_n$ converges. Assuming that $s_n$ converges then $lim_ntoinfty s_n=lim_ntoinfty s_n-1$ so $lim_ntoinfty (s_n-s_n-1)=0$ however $$s_n-s_n-1=a_1+a_2+cdots +a_n-(a_1+a_2+cdots+a_n-1)=a_n$$
So it must be $lim_ntoinfty a_n=0$ otherwise the sum doesn't converge. However if $lim_ntoinftya_n = 0$ that doesn't mean that the series converge but if $lim_ntoinftya_n = L$ and $Lneq 0$ then it diverges.
answered 35 mins ago
kingW3
10.8k72553
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up vote
0
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You may consider partial sums and split them up:
$$sum_i=1^n fracii+1 = sum_i=1^n left(1 - frac1i+1 right) = n - sum_i=1^n frac1i+1$$ $$stackrelsum_i=1^n frac1i+1 leq ln ngeq n - ln n = n(1- fracln nn) stackreln to infty longrightarrow infty$$
answered 41 mins ago
trancelocation
5,8801516
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up vote
0
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note:
$$sum_r=1^inftyfracrr+1=sum_r=1^infty1-sum_r=1^inftyfrac1r+1$$
neither of these series converge
answered 17 mins ago
Henry Lee
87410
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Ok, let's start from the very beginning, to try to clarify your doubts.
The writing $sum_i=1^inftyfracii+1$ doesn't mean that you are actually performing a sum of an infinite number of elements. It's a notation which stands for
$$lim_nrightarrowinftysum_i=1^nfracii+1.$$
This means that you are creating a sequence $sum_i=1^nfracii+1_n=1,2,...$
called sequence of partial sums and you are studying its limit for $n$ approaching $infty$. The series $sum_i=1^inftyfracii+1$ is said convergent if the sequence of partial sums converges.
Now when you say 'if I take the limit it is $1$' you are not referring to the limit of the parial sums. You are talking about $lim_irightarrowinftyfracii+1$. But why?
Well, this is because you have a result which tells you that if you have a series of the form
$sum_i=1^inftya_i$ which converges, then $lim_irightarrowinftya_i=0$.
In your case $a_i=fracii+1$ and you have $lim_irightarrowinftyfracii+1=1$. This means that your series is not convergent. As explained in the other answers the sequence of parial sums is going to $infty$ and the point is exactly that $fracii+1$ is approaching $1$.
Now the second part. Having $lim_irightarrowinftya_i=0$ is not enough to guaratee that the series $sum_i=1^infty a_i$ is convergent. And the standard example is the one you considered in one of the comments, namely $sum_i=1^inftyfrac1i$. For this series you have $lim_irightarrowinftyfrac1i=0$ but nonetheless the series is divergent.
answered 39 mins ago
Uskebasi
9021417
add a comment |Â
up vote
2
down vote
accepted
Ok, let's start from the very beginning, to try to clarify your doubts.
The writing $sum_i=1^inftyfracii+1$ doesn't mean that you are actually performing a sum of an infinite number of elements. It's a notation which stands for
$$lim_nrightarrowinftysum_i=1^nfracii+1.$$
This means that you are creating a sequence $sum_i=1^nfracii+1_n=1,2,...$
called sequence of partial sums and you are studying its limit for $n$ approaching $infty$. The series $sum_i=1^inftyfracii+1$ is said convergent if the sequence of partial sums converges.
Now when you say 'if I take the limit it is $1$' you are not referring to the limit of the parial sums. You are talking about $lim_irightarrowinftyfracii+1$. But why?
Well, this is because you have a result which tells you that if you have a series of the form
$sum_i=1^inftya_i$ which converges, then $lim_irightarrowinftya_i=0$.
In your case $a_i=fracii+1$ and you have $lim_irightarrowinftyfracii+1=1$. This means that your series is not convergent. As explained in the other answers the sequence of parial sums is going to $infty$ and the point is exactly that $fracii+1$ is approaching $1$.
Now the second part. Having $lim_irightarrowinftya_i=0$ is not enough to guaratee that the series $sum_i=1^infty a_i$ is convergent. And the standard example is the one you considered in one of the comments, namely $sum_i=1^inftyfrac1i$. For this series you have $lim_irightarrowinftyfrac1i=0$ but nonetheless the series is divergent.
answered 39 mins ago
Uskebasi
9021417
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Ok, let's start from the very beginning, to try to clarify your doubts.
The writing $sum_i=1^inftyfracii+1$ doesn't mean that you are actually performing a sum of an infinite number of elements. It's a notation which stands for
$$lim_nrightarrowinftysum_i=1^nfracii+1.$$
This means that you are creating a sequence $sum_i=1^nfracii+1_n=1,2,...$
called sequence of partial sums and you are studying its limit for $n$ approaching $infty$. The series $sum_i=1^inftyfracii+1$ is said convergent if the sequence of partial sums converges.
Now when you say 'if I take the limit it is $1$' you are not referring to the limit of the parial sums. You are talking about $lim_irightarrowinftyfracii+1$. But why?
Well, this is because you have a result which tells you that if you have a series of the form
$sum_i=1^inftya_i$ which converges, then $lim_irightarrowinftya_i=0$.
In your case $a_i=fracii+1$ and you have $lim_irightarrowinftyfracii+1=1$. This means that your series is not convergent. As explained in the other answers the sequence of parial sums is going to $infty$ and the point is exactly that $fracii+1$ is approaching $1$.
Now the second part. Having $lim_irightarrowinftya_i=0$ is not enough to guaratee that the series $sum_i=1^infty a_i$ is convergent. And the standard example is the one you considered in one of the comments, namely $sum_i=1^inftyfrac1i$. For this series you have $lim_irightarrowinftyfrac1i=0$ but nonetheless the series is divergent.
answered 39 mins ago
Uskebasi
9021417
Ok, let's start from the very beginning, to try to clarify your doubts.
The writing $sum_i=1^inftyfracii+1$ doesn't mean that you are actually performing a sum of an infinite number of elements. It's a notation which stands for
$$lim_nrightarrowinftysum_i=1^nfracii+1.$$
This means that you are creating a sequence $sum_i=1^nfracii+1_n=1,2,...$
called sequence of partial sums and you are studying its limit for $n$ approaching $infty$. The series $sum_i=1^inftyfracii+1$ is said convergent if the sequence of partial sums converges.
Now when you say 'if I take the limit it is $1$' you are not referring to the limit of the parial sums. You are talking about $lim_irightarrowinftyfracii+1$. But why?
Well, this is because you have a result which tells you that if you have a series of the form
$sum_i=1^inftya_i$ which converges, then $lim_irightarrowinftya_i=0$.
In your case $a_i=fracii+1$ and you have $lim_irightarrowinftyfracii+1=1$. This means that your series is not convergent. As explained in the other answers the sequence of parial sums is going to $infty$ and the point is exactly that $fracii+1$ is approaching $1$.
Now the second part. Having $lim_irightarrowinftya_i=0$ is not enough to guaratee that the series $sum_i=1^infty a_i$ is convergent. And the standard example is the one you considered in one of the comments, namely $sum_i=1^inftyfrac1i$. For this series you have $lim_irightarrowinftyfrac1i=0$ but nonetheless the series is divergent.
answered 39 mins ago
Uskebasi
9021417
answered 39 mins ago
Uskebasi
9021417
answered 39 mins ago
Uskebasi
9021417
answered 39 mins ago
Uskebasi
9021417
9021417
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up vote
2
down vote
A series is a sequence of partial sums. In your case,
$$frac12,frac12+frac23,frac12+frac23+frac34,frac12+frac23+frac34+frac45,cdots$$
or
$$frac12,frac76,frac2312,frac16360,cdots$$
Does this sequence converge ?
Hint:
$$frac ii+1gefrac12implies S_n>frac n2$$
answered 55 mins ago
Yves Daoust
115k666209
add a comment |Â
up vote
2
down vote
A series is a sequence of partial sums. In your case,
$$frac12,frac12+frac23,frac12+frac23+frac34,frac12+frac23+frac34+frac45,cdots$$
or
$$frac12,frac76,frac2312,frac16360,cdots$$
Does this sequence converge ?
Hint:
$$frac ii+1gefrac12implies S_n>frac n2$$
answered 55 mins ago
Yves Daoust
115k666209
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A series is a sequence of partial sums. In your case,
$$frac12,frac12+frac23,frac12+frac23+frac34,frac12+frac23+frac34+frac45,cdots$$
or
$$frac12,frac76,frac2312,frac16360,cdots$$
Does this sequence converge ?
Hint:
$$frac ii+1gefrac12implies S_n>frac n2$$
answered 55 mins ago
Yves Daoust
115k666209
A series is a sequence of partial sums. In your case,
$$frac12,frac12+frac23,frac12+frac23+frac34,frac12+frac23+frac34+frac45,cdots$$
or
$$frac12,frac76,frac2312,frac16360,cdots$$
Does this sequence converge ?
Hint:
$$frac ii+1gefrac12implies S_n>frac n2$$
answered 55 mins ago
Yves Daoust
115k666209
edited 50 mins ago
answered 55 mins ago
Yves Daoust
115k666209
answered 55 mins ago
Yves Daoust
115k666209
answered 55 mins ago
Yves Daoust
115k666209
115k666209
add a comment |Â
add a comment |Â
up vote
1
down vote
It means that the sum diverges. You are summing numbers that are larger than $frac12$, meaning that the sum of the first $n$ terms will be at least $fracn2$. That is, the partial sums grow without bound. As the sum of an infinite series is defined to be the limit of the partial sums, the series diverges.
answered 1 hour ago
Theo Bendit
13.7k12045
If this series did not diverge, is there a way to find the sum of that series? For example if the series of $sum_i=0^infty frac1n$ then the sum is 0, but is there a way to have a sum other than 0?
– Sam
1 hour ago
I'm not sure what you mean. The series does diverge, and any method that begins by assuming the series converges will arrive at totally erroneous conclusions.
– Theo Bendit
56 mins ago
add a comment |Â
up vote
1
down vote
It means that the sum diverges. You are summing numbers that are larger than $frac12$, meaning that the sum of the first $n$ terms will be at least $fracn2$. That is, the partial sums grow without bound. As the sum of an infinite series is defined to be the limit of the partial sums, the series diverges.
answered 1 hour ago
Theo Bendit
13.7k12045
If this series did not diverge, is there a way to find the sum of that series? For example if the series of $sum_i=0^infty frac1n$ then the sum is 0, but is there a way to have a sum other than 0?
– Sam
1 hour ago
I'm not sure what you mean. The series does diverge, and any method that begins by assuming the series converges will arrive at totally erroneous conclusions.
– Theo Bendit
56 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It means that the sum diverges. You are summing numbers that are larger than $frac12$, meaning that the sum of the first $n$ terms will be at least $fracn2$. That is, the partial sums grow without bound. As the sum of an infinite series is defined to be the limit of the partial sums, the series diverges.
answered 1 hour ago
Theo Bendit
13.7k12045
It means that the sum diverges. You are summing numbers that are larger than $frac12$, meaning that the sum of the first $n$ terms will be at least $fracn2$. That is, the partial sums grow without bound. As the sum of an infinite series is defined to be the limit of the partial sums, the series diverges.
answered 1 hour ago
Theo Bendit
13.7k12045
answered 1 hour ago
Theo Bendit
13.7k12045
answered 1 hour ago
Theo Bendit
13.7k12045
answered 1 hour ago
Theo Bendit
13.7k12045
13.7k12045
If this series did not diverge, is there a way to find the sum of that series? For example if the series of $sum_i=0^infty frac1n$ then the sum is 0, but is there a way to have a sum other than 0?
– Sam
1 hour ago
I'm not sure what you mean. The series does diverge, and any method that begins by assuming the series converges will arrive at totally erroneous conclusions.
– Theo Bendit
56 mins ago
add a comment |Â
If this series did not diverge, is there a way to find the sum of that series? For example if the series of $sum_i=0^infty frac1n$ then the sum is 0, but is there a way to have a sum other than 0?
– Sam
1 hour ago
I'm not sure what you mean. The series does diverge, and any method that begins by assuming the series converges will arrive at totally erroneous conclusions.
– Theo Bendit
56 mins ago
If this series did not diverge, is there a way to find the sum of that series? For example if the series of $sum_i=0^infty frac1n$ then the sum is 0, but is there a way to have a sum other than 0?
– Sam
1 hour ago
If this series did not diverge, is there a way to find the sum of that series? For example if the series of $sum_i=0^infty frac1n$ then the sum is 0, but is there a way to have a sum other than 0?
– Sam
1 hour ago
I'm not sure what you mean. The series does diverge, and any method that begins by assuming the series converges will arrive at totally erroneous conclusions.
– Theo Bendit
56 mins ago
I'm not sure what you mean. The series does diverge, and any method that begins by assuming the series converges will arrive at totally erroneous conclusions.
– Theo Bendit
56 mins ago
add a comment |Â
up vote
1
down vote
You seem to be confusing the sequence $$a_i = frac ii+1$$
And the sequence of partial sums
$$s_n=sum_i=1^na_i$$
Now $$sum_i=1^infty fracii+1$$
converges iff $s_n$ converges. Assuming that $s_n$ converges then $lim_ntoinfty s_n=lim_ntoinfty s_n-1$ so $lim_ntoinfty (s_n-s_n-1)=0$ however $$s_n-s_n-1=a_1+a_2+cdots +a_n-(a_1+a_2+cdots+a_n-1)=a_n$$
So it must be $lim_ntoinfty a_n=0$ otherwise the sum doesn't converge. However if $lim_ntoinftya_n = 0$ that doesn't mean that the series converge but if $lim_ntoinftya_n = L$ and $Lneq 0$ then it diverges.
answered 35 mins ago
kingW3
10.8k72553
add a comment |Â
up vote
1
down vote
You seem to be confusing the sequence $$a_i = frac ii+1$$
And the sequence of partial sums
$$s_n=sum_i=1^na_i$$
Now $$sum_i=1^infty fracii+1$$
converges iff $s_n$ converges. Assuming that $s_n$ converges then $lim_ntoinfty s_n=lim_ntoinfty s_n-1$ so $lim_ntoinfty (s_n-s_n-1)=0$ however $$s_n-s_n-1=a_1+a_2+cdots +a_n-(a_1+a_2+cdots+a_n-1)=a_n$$
So it must be $lim_ntoinfty a_n=0$ otherwise the sum doesn't converge. However if $lim_ntoinftya_n = 0$ that doesn't mean that the series converge but if $lim_ntoinftya_n = L$ and $Lneq 0$ then it diverges.
answered 35 mins ago
kingW3
10.8k72553
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You seem to be confusing the sequence $$a_i = frac ii+1$$
And the sequence of partial sums
$$s_n=sum_i=1^na_i$$
Now $$sum_i=1^infty fracii+1$$
converges iff $s_n$ converges. Assuming that $s_n$ converges then $lim_ntoinfty s_n=lim_ntoinfty s_n-1$ so $lim_ntoinfty (s_n-s_n-1)=0$ however $$s_n-s_n-1=a_1+a_2+cdots +a_n-(a_1+a_2+cdots+a_n-1)=a_n$$
So it must be $lim_ntoinfty a_n=0$ otherwise the sum doesn't converge. However if $lim_ntoinftya_n = 0$ that doesn't mean that the series converge but if $lim_ntoinftya_n = L$ and $Lneq 0$ then it diverges.
answered 35 mins ago
kingW3
10.8k72553
You seem to be confusing the sequence $$a_i = frac ii+1$$
And the sequence of partial sums
$$s_n=sum_i=1^na_i$$
Now $$sum_i=1^infty fracii+1$$
converges iff $s_n$ converges. Assuming that $s_n$ converges then $lim_ntoinfty s_n=lim_ntoinfty s_n-1$ so $lim_ntoinfty (s_n-s_n-1)=0$ however $$s_n-s_n-1=a_1+a_2+cdots +a_n-(a_1+a_2+cdots+a_n-1)=a_n$$
So it must be $lim_ntoinfty a_n=0$ otherwise the sum doesn't converge. However if $lim_ntoinftya_n = 0$ that doesn't mean that the series converge but if $lim_ntoinftya_n = L$ and $Lneq 0$ then it diverges.
answered 35 mins ago
kingW3
10.8k72553
answered 35 mins ago
kingW3
10.8k72553
answered 35 mins ago
kingW3
10.8k72553
answered 35 mins ago
kingW3
10.8k72553
10.8k72553
add a comment |Â
add a comment |Â
up vote
0
down vote
You may consider partial sums and split them up:
$$sum_i=1^n fracii+1 = sum_i=1^n left(1 - frac1i+1 right) = n - sum_i=1^n frac1i+1$$ $$stackrelsum_i=1^n frac1i+1 leq ln ngeq n - ln n = n(1- fracln nn) stackreln to infty longrightarrow infty$$
answered 41 mins ago
trancelocation
5,8801516
add a comment |Â
up vote
0
down vote
You may consider partial sums and split them up:
$$sum_i=1^n fracii+1 = sum_i=1^n left(1 - frac1i+1 right) = n - sum_i=1^n frac1i+1$$ $$stackrelsum_i=1^n frac1i+1 leq ln ngeq n - ln n = n(1- fracln nn) stackreln to infty longrightarrow infty$$
answered 41 mins ago
trancelocation
5,8801516
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You may consider partial sums and split them up:
$$sum_i=1^n fracii+1 = sum_i=1^n left(1 - frac1i+1 right) = n - sum_i=1^n frac1i+1$$ $$stackrelsum_i=1^n frac1i+1 leq ln ngeq n - ln n = n(1- fracln nn) stackreln to infty longrightarrow infty$$
answered 41 mins ago
trancelocation
5,8801516
You may consider partial sums and split them up:
$$sum_i=1^n fracii+1 = sum_i=1^n left(1 - frac1i+1 right) = n - sum_i=1^n frac1i+1$$ $$stackrelsum_i=1^n frac1i+1 leq ln ngeq n - ln n = n(1- fracln nn) stackreln to infty longrightarrow infty$$
answered 41 mins ago
trancelocation
5,8801516
answered 41 mins ago
trancelocation
5,8801516
answered 41 mins ago
trancelocation
5,8801516
answered 41 mins ago
trancelocation
5,8801516
5,8801516
add a comment |Â
add a comment |Â
up vote
0
down vote
note:
$$sum_r=1^inftyfracrr+1=sum_r=1^infty1-sum_r=1^inftyfrac1r+1$$
neither of these series converge
answered 17 mins ago
Henry Lee
87410
add a comment |Â
up vote
0
down vote
note:
$$sum_r=1^inftyfracrr+1=sum_r=1^infty1-sum_r=1^inftyfrac1r+1$$
neither of these series converge
answered 17 mins ago
Henry Lee
87410
add a comment |Â
up vote
0
down vote
up vote
0
down vote
note:
$$sum_r=1^inftyfracrr+1=sum_r=1^infty1-sum_r=1^inftyfrac1r+1$$
neither of these series converge
answered 17 mins ago
Henry Lee
87410
note:
$$sum_r=1^inftyfracrr+1=sum_r=1^infty1-sum_r=1^inftyfrac1r+1$$
neither of these series converge
answered 17 mins ago
Henry Lee
87410
answered 17 mins ago
Henry Lee
87410
answered 17 mins ago
Henry Lee
87410
answered 17 mins ago
Henry Lee
87410
87410
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