<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] Too long for a comment.
Consider 
$$a_k=\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ and you want to analyze $\frac{a_{k+1}}{a_k}$. It is convenient to consider first
$$\log\left(\frac{a_{k+1}}{a_k}\right)=\log\left({a_{k+1}}\right)-\log\left({a_{k}}\right)$$ where 
$$\log\left({a_{k}}\right)=k^2 \log \left(\frac{k+1}{k}\right)-k\log (3)$$ making
$$\log\left(\frac{a_{k+1}}{a_k}\right)=(k+1)^2
   \log \left(\frac{k+2}{k+1}\right)-k^2 \log \left(\frac{k+1}{k}\right)+(k+1)\log(3)-k \log (3)$$ Now, use Taylor series for large values of $k$ to get
$$\log\left(\frac{a_{k+1}}{a_k}\right)=\log \left(\frac{e}{3}\right)-\frac{1}{3
   k^2}+O\left(\frac{1}{k^3}\right)$$ Take now the exponential of both sides and continue with Taylor and get
$$\frac{a_{k+1}}{a_k}=\frac{e}{3}-\frac{e}{9 k^2}+O\left(\frac{1}{k^3}\right)$$
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