5
$\begingroup$

What criteria can I use to prove the convergence of $$ \sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?} $$

My idea was to use ratio test: $$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$ $$\displaystyle\frac{2}{\left(\, n + 1\,\right)^{2} + 2\left(\, n + 1\,\right)} \frac{n^{2} + 2n}{2} = \frac{n^{2} + 2n}{n^{2} + 4n + 3}$$

Of course $\displaystyle n^{2} + 2n \lt n^{2} + 4n + 3$ for all $\displaystyle n$ , but $$\displaystyle\lim \limits_{n \to \infty} \frac{n^{2} + 2n}{n^{2} + 4n + 3}=1$$ so I am not quite sure if I can apply ratio test.

$\endgroup$
4
$\begingroup$

You have $\frac{2}{n^2+n} \le \frac{2}{n^2}$. Thus you can use the direct comparision test to prove the convergence (if you already have proven in your course that $\sum_{n=1}^\infty \frac 1{n^2}$ and thus also $\sum_{n=1}^\infty \frac 2{n^2}$ converges).

Answer to your 2nd question: You never can apply the ratio test if the limit is 1 (as in this case).

$\endgroup$
8
$\begingroup$

I would write out the first few terms and see how they cancel.

$\endgroup$
  • $\begingroup$ of course; but to do this I have to prove absolute convergence first $\endgroup$ – Christian Dec 11 '14 at 18:42
  • 2
    $\begingroup$ @Christian No, you do not. $\endgroup$ – Andrés E. Caicedo Dec 11 '14 at 18:45
  • $\begingroup$ @Christian You can look at the partial sums and see how they cancel without worrying about absolute convergence. Your problem would be with $\sum \limits_{n=1}^{\infty} \frac1n - \sum \limits_{m=1}^{\infty}\frac{1}{m+2}$ $\endgroup$ – Henry Dec 11 '14 at 18:47
  • 1
    $\begingroup$ It might be useful to mention Telescoping Series. $\endgroup$ – robjohn Dec 16 '14 at 13:04
  • $\begingroup$ The reason that doing this works is that it allows you to calculate the partial sums directly, which you can use to show convergence of the series by definition (i.e. showing that the partial sums converge) $\endgroup$ – Joshua Mundinger Dec 16 '14 at 16:44
7
$\begingroup$

How about computing the partial sums? For any $N>2$ we have:

$$\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{3}{2}-\frac{1}{N+1}-\frac{1}{N+2}$$ hence: $$\left|\frac{3}{2}-\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right)\right|\leq\frac{2}{N}$$ ensures convergence (towards $\frac{3}{2}$).

$\endgroup$
2
$\begingroup$

Here's another way to prove the convergence $$ \sum \limits_{n=1}^{\infty} \left(\frac1n - \frac{1}{n+2}\right)= \sum \limits_{n=1}^{\infty} \frac{n+2-n}{n(n+2)} $$ $$= \sum \limits_{n=1}^{\infty} \frac{2}{n^2+2n}=2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} $$ Also note that $$ \left|\frac{1}{n^2+2n}\right|\leq \left|\frac{1}{n^2}\right| $$ And by the p-series test, we have $$ \sum \limits_{n=1}^{\infty} \left|\frac{1}{n^2}\right| \Rightarrow \mbox{converges} $$ Which implies that $$ \sum \limits_{n=1}^{\infty} \frac{1}{n^2} \Rightarrow \mbox{converges absolutely} $$ Therefore, by the direct comparison test $$ 2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} \Rightarrow \mbox{converges absolutely} $$ Absolute convergence implies convergence. Also a convergent series multiplied by $2$ is still a convergent series.

$\endgroup$
  • 1
    $\begingroup$ I hope the downvoter revisits this answer. I don't think it deserves a downvote, especially after the last edit. (+1) $\endgroup$ – robjohn Dec 16 '14 at 13:02
1
$\begingroup$

You can use the limit comparison test. Let $a_{n} = \frac{2}{n^{2}+2n}$ and consider $b_{n} = \frac{1}{n^{2}}$ then

$$\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}}= \lim_{n\rightarrow \infty}\frac{2n^{2}}{n^{2}+2n} = 2$$

hence $\sum a_{n}$ and $\sum b_{n}$ converge or diverge together.. but $b_{n}$ is a convergent $p$-series.

That being said, my up vote is for Mark Bennet.

$\endgroup$
0
$\begingroup$

How about the comparison test?

$\endgroup$
  • $\begingroup$ with $\frac{1}{n^2}$ ? $\endgroup$ – Christian Dec 11 '14 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.