I need to try and solve the following problem, as stated in the title:
For what $x's$ does $\sum_{k=1}^{\infty} \frac{(k+1)^{k^2}}{k^{k^2+2}} x^k$ converge?
Attempt:
First of all, we try to find the radius of convergence for our given sum. Let $a_k := \frac{(k+1)^{k^2}}{k^{k^2+2}}$. The radius of convergence is given by:
$$ R = (\limsup_{k\rightarrow \infty} |a_k|^{1/k})^{-1}$$
Plugging our value for $a_k$ in the expression above, we have:
$$ R = \left (\limsup_{k\rightarrow \infty} \left |\frac{(k+1)^{k^2}}{k^{k^2+2}}\right |^{1/k}\right)^{-1}$$
Which can be simplified to: $$R = \left (\limsup_{k\rightarrow \infty} \left |\frac{(k+1)^{k}}{k^{k+2/k}}\right |\right)^{-1}$$
Examining the numerator and the denominator separatly, we have that:
$(k+1)^{k} = e^{k\ln(k+1)}$ and $k^{k+2/k} = e^{(k+2/k)\ln(k)}$
Hence:
$$ R = \left (\limsup_{k\rightarrow \infty} \left |e^{k\ln(k+1)-(k+2/k)\ln(k)}\right |\right)^{-1}$$
$$ R = \left (\limsup_{k\rightarrow \infty} \left |e^{k\ln(1+1/k)-2\ln(k)/k}\right |\right)^{-1}$$
Therefore $R = 1/e$. So, $\forall |x| < 1/e$, we have absolute convergence. Hence, convergence too. We now need to examine the cases $|x| = 1/e$ separately:
Case $x = 1/e$
I then have:
$\sum_{k=1}^{\infty} \frac{(k+1)^{k^2}}{e^k k^{k^2+2}}$.
However, I've a hard time determining on how to show that this is either convergent / divergent, since there're so many complicated factors involved. I'd be glad if you could share any tips on how to proceed with this.
Thanks!
