Prove $\sum_{n=2}^{\infty}\ln\left(1+\frac{(-1)^n}{n^p}\right)$ converges if and only if $p>\frac 12$

Question

I am trying to prove that the following sequence converges: $$\sum_{n=2}^{\infty}\ln\left(1+\frac{(-1)^n}{n^p}\right)$$ if and only if $p>\frac 12$.

I've seen solutions to this exact problem here, but I am not looking for a general solution. I've tried to solve this problem, and could not continue my solution, so I came here to ask for your help on how to continue.

My solution

$(\star)$ I have proven that given a sequence $a_n$, such that $\lim_{n\to\infty}a_n=0$, if: $\sum_{n=1}^{\infty}(a_{2n}+a_{2n+1})$ converges or diverges to infinity, then $\sum_{n=2}^{\infty}a_n$ converges or diverges to infinity, respectively.

I have also proven that $\sum_{n=2}^{\infty}\ln\left(1+\frac{(-1)^n}{n^p}\right)$ converges absolutely for every $p>1$, converges for $p=1$, and diverges for $p=\frac 12$. So what I have left, essentially, is to prove that the series converges for every $\frac 12<p<1.$

We can see that:

$$\sum_{n=1}^{\infty}(a_{2n}+a_{2n+1})\equiv\sum_{n=1}^{\infty}\left(\ln(1+\frac{1}{(2n)^p})+\ln(1-\frac{1}{(2n+1)^p})\right)$$

Since the sequence is negative, we can use the limit test with (after using logarithm rules):

$$\frac{(2n)^p-(2n+1)^p+1}{(4n^2+2n)^p}$$

Now, on the one hand, I don't have the logarithm anymore; But on the other hand, I don't know how to deal with this series. I tried to use the limit test again with $\frac{1}{n^{2p}}$, but to no avail.

I would be very glad to hear how to continue my solution, or rather simplify it. I prefer using the claim I've proven (marked with $(\star)$).

Thank you very much!

Answer 1

Your limit test can be made to work. $$ \lim_{x\rightarrow\infty}\frac{\frac{(2x)^p-(2x+1)^p+1}{(4x^2+2x)^p}}{\frac{1}{x^{2p}}}=\lim_{x\rightarrow\infty}\frac{(2x)^p-(2x+1)^p+1}{(4+\frac{2}{x})^p}=\frac{1}{4^p} $$ Here I used the fact that $\lim_{x\rightarrow\infty}((2x+1)^p-(2x)^p)=0$. You can quickly prove this by noting that $f:x\rightarrow x^p$ is a concave function for $p<1$ ($f''<0$), so you can say that $(2x+1)^p-(2x)^p < 2p(2x)^{p-1}$. But $p-1<0$, so this last expression goes to $0$.

Answer 2

From the generalized Bernoulli inequality you have for $0<p\le1$: $$ \big|(2n)^p-(2n+1)^p+1\big| \le 1+ (2n)^p \big((1+\frac{1}{2n})^p- 1\big) \le 1+ (2n)^p\frac{p}{2n}$$ You have then $$ \left|\frac{(2n)^p-(2n+1)^p+1}{(4n^2 + 2n)^p}\right| \le \frac{1+ p(2n)^{p-1}}{(2n)^{2p}(1+\frac{1}{2n})^p} $$ $(2n)^{p-1} \rightarrow 0$, so by comparison test, the series is convergent for $p>\frac12$.