How to test the convergence for $\sum _{n=3}^{\infty }\:\left(-\frac{1}{\sqrt{ln\left(n\right)}}\right)^{n+1}$? How to test the convergence for
$$\sum _{n=3}^{\infty }\:\left(-\frac{1}{\sqrt{\ln\left(n\right)}}\right)^{n+1}$$
Can I use the comparison test for this one?
For example:
$n≥\ln(n)$
$\sqrt{n}\ge \sqrt{\ln\left(n\right)}$
$-\sqrt{n}\le -\ln\left(n\right)$
$\frac{1}{-\sqrt{n}}\ge \frac{1}{-\sqrt{\ln\left(n\right)}}$
and then I use the the p series for p = 1 and say that this diverges? or can I only use this inequality for convergent?
Thanks :)
 A: Hint: After rewriting it as
$$\sum _{n=3}^{\infty }\:\left(-\frac{1}{\sqrt{\ln\left(n\right)}}\right)^{n+1} = \sum _{n=3}^{\infty }\: (-1)^{n+1}\left(\frac{1}{\sqrt{\ln\left(n\right)}}\right)^{n+1} = \sum_{n=3}^\infty (-1)^{n+1} a_n$$
notice that it is an alternating series and luckily $a_n\to0$ decreasing monotonically as $n \to \infty$.
A: For large enough $n$, $|\frac{-1}{\sqrt(ln(n))}| \leq \frac{1}{2}$. So as $n$ increases its absolutely bounded by a geometric series. So by comparison of series ur series converges absolutely.
A: As @Vivid did, consider
$$a_n=\left(\frac{1}{\sqrt{\log\left(n\right)}}\right)^{n+1}\implies \log(a_n)=-\frac 12(n+1) \log (\log (n))$$
$$\log(a_{n+1})-\log(a_n)=\frac{1}{2} (n+1) \log (\log (n))-\frac{1}{2} (n+2) \log (\log (n+1))$$ Using Taylor series
$$\log(a_{n+1})-\log(a_n)=-\frac{1}{2} \log (\log (n))-\frac{1}{2 \log (n)}+\cdots$$
$$\frac{a_{n+1} }{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac{e^{-\frac{1}{2 \log (n)}}}{\sqrt{\log (n)}}\quad \to ~~0$$
A: \begin{equation}
\begin{split}
\lim _{n\to \infty }\left(\frac{\left(-\frac{1}{\sqrt{\ln \left(n+1\right)}}\right)^{n+2}}{\left(-\frac{1}{\sqrt{\ln \left(n\right)}}\right)^{n+1}}\right)
&=\lim _{n\to \infty }\left(-\ln ^{\frac{1+n}{2}}\left(n\right)\cdot \ln ^{-1-\frac{n}{2}}\left(1+n\right)\right)\\
&=-\lim _{n\to \infty }\left(\ln ^{\frac{1+n}{2}}\left(n\right)\cdot \ln ^{-1-\frac{n}{2}}\left(1+n\right)\right)\\
&=-\lim _{n\to \infty }\left(\sqrt{\ln ^{n+1}\left(n\right)\cdot \ln ^{-n-2}\left(1+n\right)}\right)\\
&=-\sqrt{\lim _{n\to \infty }\left(\ln ^{n+1}\left(n\right)\cdot \ln ^{-n-2}\left(1+n\right)\right)}\\
&=-\sqrt{\lim _{n\to \infty }\left(\exp \left(\ln ^2\left(n\right)\left(n+1\right)+\ln ^2\left(n+1\right)\left(-n-2\right)\right)\right)}\\
&=-\sqrt{\exp{\lim _{n\to \infty }\left(\ln ^2\left(n\right)\left(n+1\right)+\ln ^2\left(n+1\right)\left(-n-2\right)\right)}}
\end{split}
\end{equation}
Since $\lim _{n\to \infty }\left(\ln ^2\left(n\right)\left(n+1\right)+\ln ^2\left(n+1\right)\left(-n-2\right)\right)$ is $-\infty$, the limit of the whole expression is $0$. Hence, convergent by ratio test.
