- #1

- 1,462

- 44

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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 1,462

- 44

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.

- #2

- 18,242

- 20,169

- #3

Science Advisor

- 8,134

- 568

##|\frac{1}{n^2-n/2}|\le \frac{2}{n^2}## will do.

- #4

- 1,462

- 44

How did you find that? I can see why it's true, but I don't see how I'd spot it: $$n^2\ge n \implies 2n^2 - n \ge n^2 \implies 2(n^2 - n/2) \ge n^2 \implies \frac{2}{n^2} \ge \frac{1}{n^2-n/2}$$.##|\frac{1}{n^2-n/2}|\le \frac{2}{n^2}## will do.

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

- 18,242

- 20,169

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.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.

- #6

Science Advisor

Gold Member

- 6,376

- 8,714

Maybe also partial fractions?

- #7

Science Advisor

Gold Member

- 1,260

- 597

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##

- - - -

Last edited:

- #8

Science Advisor

Gold Member

- 6,376

- 8,714

Kind of weird what you get here:Maybe also partial fractions?

## 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

- 18,242

- 20,169

But we do not know, whether this is still true with the ##2## in the nominator of all odd indices.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.

- #10

Science Advisor

Gold Member

- 6,376

- 8,714

Why not, this is just a rescaling. If the sum is finite, then so is twice the sum., isn't it?But we do not know, whether this is still true with the ##2## in the nominator of all odd indices.

- #11

Science Advisor

Gold Member

- 6,376

- 8,714

Just rewrite:But we do not know, whether this is still true with the ##2## in the nominator of all odd indices.

##\Sigma 4( \frac {1}{2n-1} - \frac {1}{4n} ) ##

- #12

Science Advisor

- 8,134

- 568

When n=1, ##\frac{1}{n^2-n/2}=2##, so ##\frac{2}{n^2}## works here and therefore for larger n.How did you find that? I can see why it's true, but I don't see how I'd spot it:

Share:

- Replies
- 5

- Views
- 255

- Replies
- 2

- Views
- 188

- Replies
- 3

- Views
- 567

- Replies
- 18

- Views
- 870

- Replies
- 1

- Views
- 986

- Replies
- 1

- Views
- 514

- Replies
- 3

- Views
- 701

- Replies
- 5

- Views
- 1K

- Replies
- 5

- Views
- 711

- Replies
- 3

- Views
- 924