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Do the following converge:

$\sum_{n=2}^\infty {\frac{n+2}{n^3-2n^2+1}}$

For this one I think the answer is no I just can't prove it. I split it up into partial fractions and got: $\frac{3n+1}{n^2-n-1}-\frac{3}{n-1}$

but after that I'm stumped :(

The second part: $\sum_{n=1}^\infty {\frac{1}{n}+\frac{(-1)^n}{n^2}}$

This one I think also diverges, I tried to use comparison test but didn't get far.

I think I am missing something obvious for both questions.

Mark Viola
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babylon
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  • The horribly informal way is to just look at the highest ordered terms on the top and bottom. $\sum\limits_{n=1}^\infty \frac{n+2}{n^3-2n^2+1}$ acts like $\sum\limits_{n=1}^\infty \frac{n}{n^3}$. With that as a hint, see if you can get an appropriate comparison. – JMoravitz Dec 15 '15 at 17:45
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    Firstly, the counter must be started at $n=2$ and not $n=1$ (because if $n=1$ then $n^3-2n^2+1=0$). Secondly, it is convergent. You can use the limit comparison test and use the convergent series $\sum_{n=2}^{n=\infty}\frac{1}{n^2}$ . – Albert Dec 15 '15 at 17:50

3 Answers3

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Say $u_n=\frac{n+2}{n^3-2n^2+1}$ and $v_n=\frac{1}{n^2}$.

Then $$\lim_{n\to\infty} \frac{u_n}{v_n} =\lim_{n\to\infty} \frac{\frac{n+2}{n^3-2n^2+1}}{\frac{1}{n^2}} $$ $$=\lim_{n\to\infty} \frac{1+\frac{2}{n}}{1-\frac{2}{n}+\frac{1}{n^3}}=1>0$$

This limit is finite and $\sum v_n$ is convergent by p-series test.

Hence by comparison test, the given series converges.

SchrodingersCat
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For the second one (which should have been a separate question) note the following:

If $\sum a_n$ diverges and $\sum b_n$ converges, then $\sum (a_n+b_n)$ diverges.

This follows directly from the observation that $\sum (a_n+b_n) = \sum a_n + \sum b_n$

Let $a_n = \frac{1}{n}$ and $b_n = \frac{(-1)^n}{n^2}$. Each of these sums should be familiar to you and can immediately give you your conclusion.

JMoravitz
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Here is a fairly simple approach. Note that for all $n\ge2$ the following inequalities hold

$$n^3-2n^2+1\ge \frac18 n^3$$

and

$$n+2\le 2n$$

Therefore, we have

$$\frac{n+2}{n^3-2n^2+1}\le\frac{2n}{\frac18 n^3}=\frac{16}{n^2}$$

Inasmuch as $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$, we can write

$$\sum_{n=2}^{\infty}\frac{n+2}{n^3-2n^2+1}\le 16\sum_{n=2}^{\infty}\frac{1}{n^2}=\frac83 \pi^2-16$$

And we are done!

Mark Viola
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  • Please let me know how I can improve my answer. I really want to give you the best answer I can. Happy Holidays! - Mark – Mark Viola Dec 29 '15 at 16:01