<s>[INST] Testing $\sum\limits_{k=1}^∞(\frac{k+1}k)^{k^2}3^{-k}$ for convergence and absolute convergence 
Test $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ for convergence and absolute convergence.

We apply the ratio test for $\displaystyle \sum_{k=1}^{\infty}\left|\left(\frac{k+1}{k}\right)^{k^2}3^{-k}\right|$:
$$
\left|\frac{\left(\dfrac{k+2}{k+1}\right)^{(k+1)^2}3^{-(k+1)}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|=\left|\frac{\left(\left(\dfrac{k+2}{k+1}\right)^{k}\right)^2\left(\dfrac{k+2}{k+1}\right)^{2k+1}3^{-k}\dfrac{1}{3}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|→\frac{e^2\cdot e^2\cdot\dfrac13}{e^2}=\frac{1}{3}e^2.
$$
Since $\dfrac{1}{3}e^2 \geq 1$, the series diverges. Is this correct? I feel like I made a mistake somewhere that I cannot pin down.
 [/INST] FYI, estimation of upper bound:
$\sum\limits_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}=\sum\limits_{k=1}^{\infty}\left(\frac{1}{k}+1\right)^{k^2}3^{-k}$
Using binominal theorem and the following unequality: $\begin{pmatrix}n \\ {k} \end{pmatrix}\lt \frac{n^k}{k!}$ we get:
$\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\begin{pmatrix}{k^2} \\ {j} \end{pmatrix} k^{-j}=\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\frac{k^{2j}}{j!} k^{-j}=\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\frac{k^{j}}{j!}\lt\sum\limits_{k=1}^{\infty}\big(\frac{e}{3}\big)^k=\frac{e}{3-e}$
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