Showing that an infinite sum converges when a sign in the denominator is flipped

  • #1
1,462
44
##\displaystyle \sum_{n=1}^\infty\frac{1}{n^2+n/2}## converges by the direct comparison test: ##\displaystyle \left|\frac{1}{n^2+n/2}\right| \le \left|\frac{1}{n^2}\right|##, and ##\displaystyle \sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}##.

But what if we want to show that ##\displaystyle \sum_{n=1}^\infty\frac{1}{n^2-n/2}## converges? It doesn't seem like we can use the direct comparison test anymore. Also, the ratio test is inconclusive.
 

Answers and Replies

  • #2
fresh_42
Mentor
Insights Author
2021 Award
16,436
15,483
You can write ##n^2-\frac{n}{2}## as ##k^2 - c > m^2## and bound the sum as some finite start sequence plus ##\sum \frac{1}{m^2}\,.##
 
  • #3
mathman
Science Advisor
8,046
535
##|\frac{1}{n^2-n/2}|\le \frac{2}{n^2}## will do.
 
  • #4
1,462
44
##|\frac{1}{n^2-n/2}|\le \frac{2}{n^2}## will do.
How did you find that? I can see why it's true, but I don't see how I'd spot it: $$n^2\ge n \implies 2n^2 - n \ge n^2 \implies 2(n^2 - n/2) \ge n^2 \implies \frac{2}{n^2} \ge \frac{1}{n^2-n/2}$$.

Basically I'm trying to make sure I know how to find upper bounds on expressions that can't obviously be bounded above. When the sign is positive is easy to see, but it doesn't seem immediate when the sign is negative.
 
  • #5
fresh_42
Mentor
Insights Author
2021 Award
16,436
15,483
Basically I'm trying to make sure I know how to find upper bounds on expressions that can't obviously be bounded above. When the sign is positive is easy to see, but it doesn't seem immediate when the sign is negative.
Of course I don't know what @mathman thought. But a closer look on the summands show that they go with ##O(\frac{1}{n^2})## and some minor changes cannot change this general behavior. Now ##O(\frac{1}{n^2})## means ##< c \cdot \frac{1}{n^2}## so it's very natural to simply increase ##c##. I would probably have chosen ##c=100## to avoid counting low values of ##n## and to emphasize, that it doesn't matter how big ##c## is as long as it is a constant - if I only had seen this possibility as fast.
 
  • Like
Likes Mr Davis 97
  • #6
WWGD
Science Advisor
Gold Member
5,854
6,791
Maybe also partial fractions?
 
  • #7
StoneTemplePython
Science Advisor
Gold Member
1,203
597
##\displaystyle \sum_{n=1}^\infty\frac{1}{n^2+n/2}## converges by the direct comparison test: ##\displaystyle \left|\frac{1}{n^2+n/2}\right| \le \left|\frac{1}{n^2}\right|##, and ##\displaystyle \sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}##.

But what if we want to show that ##\displaystyle \sum_{n=1}^\infty\frac{1}{n^2-n/2}## converges? It doesn't seem like we can use the direct comparison test anymore. Also, the ratio test is inconclusive.

Another idea is to just bound the tails of your series (i.e. you can always ignore finitely many terms for convergence questions). So
##(n-k)^2 \lt n^2 - \frac{n}{2}##

for ##n## large enough (i.e. all ##n \geq N##) and suitable choice of natural number ##k \lt N##. You can choose exact values if you desire by expanding to get

##(n-k)^2 = n^2 - 2nk + k^2 \lt n^2 - \frac{n}{2}##
re-arranging gives
## 0\lt -k^2 + 2nk - \frac{n}{2} = -k^2 + \frac{4nk -n}{2}= -k^2 + \frac{n(4k -1)}{2} ##

qualitatively you need to recognize that ##k## is fixed (you choose it) and ##n## monotonically increases the value of the RHS, so it is easily positive (and easy to find an ##N## if you must).

now take advantage of positivity, invert
##(n-k)^2 \lt n^2 - \frac{n}{2}##
to get

##0\lt \frac{1}{n^2 - \frac{n}{2}}\lt \frac{1}{(n-k)^2}##

summing over the bound:

$$0 \lt \sum_{n\geq N} \frac{1}{n^2 - \frac{n}{2}}\leq \sum_{n\geq N} \frac{1}{(n-k)^2}= \sum_{n\geq (N-k)} \frac{1}{n^2} \leq \sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{6} \lt \infty $$

where the RHS comes from recognizing that ##k## exists to downshift your summation.

- - - -
in general repeat for
##(n-k)^2 \leq \beta n^2 + \alpha n + c##
- - - -

edit: this may be close to what Fresh had in mind for post ##2##
 
Last edited:
  • #8
WWGD
Science Advisor
Gold Member
5,854
6,791
Maybe also partial fractions?
Kind of weird what you get here:

## 2( \frac {2}{2n-1}- \frac {1}{n}) ## , so, since this was shown to converge, it seems we get that the sum of the odd terms in the harmonic series minus the even ones converges. But we knew this, that the alternating sum ## \Sigma (-1)^n/n ## converges.
 
  • #9
fresh_42
Mentor
Insights Author
2021 Award
16,436
15,483
Kind of weird what you get here:

## 2( \frac {2}{2n-1}- \frac {1}{n}) ## , so, since this was shown to converge, it seems we get that the sum of the odd terms in the harmonic series minus the even ones converges. But we knew this, that the alternating sum ## \Sigma (-1)^n/n ## converges.
But we do not know, whether this is still true with the ##2## in the nominator of all odd indices.
 
  • #10
WWGD
Science Advisor
Gold Member
5,854
6,791
But we do not know, whether this is still true with the ##2## in the nominator of all odd indices.
Why not, this is just a rescaling. If the sum is finite, then so is twice the sum., isn't it?
 
  • #11
WWGD
Science Advisor
Gold Member
5,854
6,791
But we do not know, whether this is still true with the ##2## in the nominator of all odd indices.
Just rewrite:

##\Sigma 4( \frac {1}{2n-1} - \frac {1}{4n} ) ##
 
  • #12
mathman
Science Advisor
8,046
535
How did you find that? I can see why it's true, but I don't see how I'd spot it:
When n=1, ##\frac{1}{n^2-n/2}=2##, so ##\frac{2}{n^2}## works here and therefore for larger n.
 

Related Threads on Showing that an infinite sum converges when a sign in the denominator is flipped

Replies
3
Views
1K
M
Replies
1
Views
5K
Replies
2
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
9
Views
8K
Replies
4
Views
1K
Replies
7
Views
3K
Replies
7
Views
5K
Replies
2
Views
6K
Top