The question considers the power series $\displaystyle\sum_{k=1}^\infty \frac{\ln^kk}{k^k}x^k$.
The question is to show that the radius of convergence is $(-\infty,\infty)$, but without using the nth root test of convergence.
[I can do the question using the nth root test with no issue.]
In lectures we have only used the usual radius of convergence formula
$R=\displaystyle \lim_{k\rightarrow \infty}\frac{a_k}{a_{k+1}}$ derived from the ratio test.
But this yields an impossible expression!
I get
$R=\displaystyle \lim_{k\rightarrow \infty}\frac{\ln^kk}{k^k}\cdot \frac{(k+1)^{k+1}}{\ln^{k+1}(k+1)}$
$\therefore R>\displaystyle \lim_{k\rightarrow \infty}\frac{\ln^kk}{k^k}\cdot \frac{k^{k+1}}{\ln^{k+1}(k+1)}$
$\therefore R>\displaystyle \lim_{k\rightarrow \infty}\frac{k\ln^kk}{\ln^{k+1}(k+1)}$
I need to show this diverges. Of course $k> \ln k$ but I can't see how to proceed here.
L'Hopital's Rule applies but the derivatives are awful!
Wolfram Alpha says it diverges!
Any hints on what I am missing and how to proceed please. Thanks 🙂
Best Answer
Note that $$\frac{k\ln^kk}{\ln^{k+1}(k+1)} =\frac k{\ln(k+1)}\cdot\left(\frac {\ln k}{\ln(k+1)}\right)^k =\frac k{\ln(k+1)}\cdot\left(1-\frac {\ln(1+\frac 1k)}{\ln (k+1)}\right)^k $$ and by Bernnoulli together with $\ln(1+x)\le x$, $$\left(1-\frac {\ln(1+\frac1 k)}{\ln (k+1)}\right)^k\ge 1- \frac {k\ln(1+\frac 1k)}{\ln (k+1)}\ge 1-\frac{1}{\ln(k+1)}.$$