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.
 A: 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
$$\frac{a_{k+1}}{a_k}=\frac{e}{3}-\frac{e}{9 k^2}+O\left(\frac{1}{k^3}\right)$$
A: Well known fact: $(1+1/k)^k < e, k=1,2,\dots$ Thus
$$(1+1/k)^{k^2}3^{-k}=[(1+1/k)^k]^k\cdot3^{-k} < e^k3^{-k} = (e/3)^k.$$
Since $0<e/3<1,$ $\sum (e/3)^k$ is a convergent geometric series. By the comparison test, $\sum (1+1/k)^{k^2}3^{-k}$ converges. (It's a positive series, so convergence and absolute convergence are the same.)
A: Yet another way:
\begin{align*}
\frac{a_{k+1}}{a_{k}} & =\frac{1}{3}\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{(1+k)^{2}}\\
 & =\frac{1}{3}\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{k^{2}+k}\left(1+\frac{1}{1+k}\right)^{k+1}.
\end{align*}
Now, if
$$
b_{k}\equiv\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{k^{2}+k}
$$
has limit one, we can conclude $a_{k+1}/a_{k}\rightarrow e/3$. Take
logs to get
$$
\log b_{k}=k\left[\left(1+k\right)\log\left(1+\frac{1}{1+k}\right)-k\log\left(1+\frac{1}{k}\right)\right].
$$
As in Claude Leibovici's answer, expand around $\infty$ to get
$$
\log b_{k}=k\left(\frac{1}{2k^{2}}+O\left(\frac{1}{k^{3}}\right)\right)\rightarrow0
$$
so that $b_{k}=e^{\log b_k}\rightarrow e^0=1$, as desired.
A: 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:
$\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\begin{pmatrix}{k^2} \\ {j} \end{pmatrix} k^{-j}=\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\frac{k^{2j}}{j!} k^{-j}=\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\frac{k^{j}}{j!}\lt\sum\limits_{k=1}^{\infty}\big(\frac{e}{3}\big)^k=\frac{e}{3-e}$
A: It has already been pointed out where you've made a mistake, so here's an alternative approach:
By the root test,
$$\lim_{n\to\infty}\sqrt[n]{\left|\left(1+\frac1n\right)^{n^2}3^{-n}\right|}=\lim_{n\to\infty}\left(1+\frac1n\right)^n3^{-1}=\frac e3<1$$
the series converges.
