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

[Mr Davis 97]

$\displaystyle \sum_{n=1}^\infty\frac{1}{n^2+n/2}$ 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.

 

[fresh_42]

You can write $n^2-\frac{n}{2}$ as $k^2 - c > m^2$ and bound the sum as some finite start sequence plus $\sum \frac{1}{m^2}\,.$

 

[mathman]

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

 

[Mr Davis 97]

$|\frac{1}{n^2-n/2}|\le \frac{2}{n^2}$ 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.

 

[fresh_42]

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.

 

[WWGD]

Maybe also partial fractions?

 

[StoneTemplePython]

$\displaystyle \sum_{n=1}^\infty\frac{1}{n^2+n/2}$ 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 $2$

 

[WWGD]

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.

 

[fresh_42]

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.

 

[WWGD]

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?

 

[WWGD]

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

Just rewrite:

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

 

[mathman]

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.