# Convergence of series $\sum_{n=1}^\infty \frac{1+x^{2n}}{n^6}$

I was asked to determine for what $x\in\mathbb{R}$ does $\displaystyle \sum_{n=1}^\infty \frac{1+x^{2n}}{n^6}$.
My attempt is:
$\displaystyle \frac{1+x^{2n}}{n^6}=\frac{1}{n^6}+\frac{x^{2n}}{n^6}$.

I was told the "two out of three" rule:

If $c_n=a_n+b_n$, then $\sum C_n=\sum a_n +\sum b_n$, provided two of the three series are known to converge.

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^6}$ converges by the $p$-test.
$\displaystyle \sum_{n=1}^\infty \frac{x^{2n}}{n^6}$ converges when $|x|\leq 1$ by the ratio test.

My professor said that I need to be very careful with the logic of the rule.
How do I show that $\sum c_n$ converges $\Leftrightarrow \sum a_n +\sum b_n$ converges or other way to show $\sum c_n$ converges?

Thank you.

For $|x|>1$, then $\dfrac{x^{2n}}{n^{6}}$ does not converge to $0$, then $\displaystyle\sum\dfrac{x^{2n}}{n^{6}}$ does not exist, if $\displaystyle\sum\dfrac{1+x^{2n}}{n^{6}}$ exists for such an $x$, then as $\displaystyle\sum\dfrac{1}{n^{6}}$ exists, then $\displaystyle\sum\dfrac{x^{2n}}{n^{6}}$ exists by standard limit rule, a contradiction.

So we conclude that for $|x|>1$, the series does not converge. And it has been argued that for $|x|\leq 1$, it converges, so the interval is then $[-1,1]$.

• Is it sufficient to say that as $\sum \frac{1}{n^6}$ converges and we want to have $\sum \frac{1+x^{2n}}{n^6}$ converge, so $\sum \frac{x^{2n}}{n^6}$ must too converge? – tooooony Mar 13 '18 at 17:29
• Because $\displaystyle\sum\dfrac{1}{n^{6}}$ converges, then $\displaystyle\sum\dfrac{1+x^{2n}}{n^{6}}$ converges if and only if $\displaystyle\sum\dfrac{x^{2n}}{n^{6}}$ converges by a standard limit rule. – user284331 Mar 13 '18 at 17:31
• Ah, limits. Thank you. – tooooony Mar 13 '18 at 17:40

For $|x|\le 1$ we have

$$0<\frac{1+x^{2n}}{n^6}\le \frac2{n^6}$$

So the series converges for $|x|\le 1$.

Now note that we have $$0<\frac{x^{2n}}{n^6}< \frac{1+x^{2n}}{n^6}$$

So, the series diverges for $|x|>1$.