For what values of p is given series convergent.
The given series is $\sum_{n=2}^{\infty} (-1)^{n-1}\left(\frac{{(\ln n)}^{p}}{n}\right)$.And the question is for what values of $p$ series converges. I try a little bit but don't know whether it is right?I try to check values of p for which series is absolutely convergent and then by using integral test I think $p<-1$.But when I use alternating series test I think $p$ will take all negative real values and $0$ also. But I am getting feeling that $p$ will take some or all positive values also as $\lim_{n\to\infty}{(\ln n)}^{p}/n=0$. Please correct me.
$\endgroup$ 41 Answer
$\begingroup$- For the absolute convergence, the series $\;\displaystyle\sum_{n\ge2}\frac{\log^p n}n=\sum_{n\ge2}\frac1{n\log^{-p}n}$ is a Bertrand's series, known to converge if and only if $-p>1$.
- For the semi-convergence, apply Leibniz' criterion for alternating series. We only have to determine for which values of $p$ $\;\dfrac{\log^p n}n$ is eventually monotonically decreasing.
So consider the function $f(x)=\dfrac{\log^p x}x$. Its derivative is $$f'(x)=\frac{p\log^{p-1}x-\log^px}{x^2}=\frac{\log^{p-1\!}x\,(p-\log x)}{x^2}$$ and, if $x>1$, it has the sign of $p-\log x$. Thus the function is decreasing if and only if $$$\log x>p\iff x >\mathrm e^{\mkern2mu p} $$ As a conclusion the sequence is eventually decreasing for all values of $p$, hence the alternating series converges for all $p$.
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