Konverge II

First we see that $\lvert x \rvert$ must be $\leq 1$ . Otherwise The nth term tends to $\infty$ as $n\to\infty$ which implies divergence .

So first we check for $x=1$ which would also be the same as checking for $x=-1$ as the power of x is even .

So for $x=1$ the series becomes :-

$\large\sum_{n=2}^{\infty} \frac{1}{n(ln(n))^{2}}$ .

Now we apply Cauchy Condensation Test .

As the $nth$ terms form a decreasing sequence we can apply Cauchy Condensation Test

So the series shares it's fate with(converges or diverges with)

$\large\sum_{n=2}^{\infty} \frac{2^n}{2^{n}(ln(2^{n}))^{2}}$

$=\large\sum_{n=2}^{\infty} \frac{1}{n^{2}(ln(2))^{2}}$

which is convergent(we can also find a closed form of this)

So series is convergent for $x=1$ and $x=-1$ .

Now let $x \in (-1,1)$ .

So the series becomes

$\large\sum_{n=2}^{\infty} \frac{x^{2n}}{n(ln(n))^{2}} \leq \sum_{n=2}^{\infty} \frac{1}{n(ln(n))^{2}}$

Now as $RHS$ converges.......we have by Direct Comparison Test . $LHS$ is also convergent.

So the series converges for any $x \in [-1,1]$

Hint: use the Ratio test, with the limit $L < 1$ . Remember to check the extremities (Integral Test).