<s>[INST] Does $\sum_{k=1}^\infty \frac{k^2}{3^k}$ converge or diverge? I am looking at the series $$\sum_{k=1}^\infty a_k $$
where $$a_k = \frac{k^2}{3^k}$$
The terms denominator grow much faster (Exponential) than the numerators (Quadratic) so I am guessing it will converge, but I also want to find the value of it since I am almost sure it does converge 
 [/INST] Your intuition is correct - the sum converges. I will show this by showing that the sum, taken from the fifth term onward, converges. This is sufficient because the first four terms are finite. 

Notice that for all $k>4$, $$2^k > k^2 \qquad \implies \qquad \frac{2^k}{3^k} > \frac{k^2}{3^k}$$
This implies $$\sum_{k=5}^\infty\, \left( \frac{2}{3}\right)^k \,>\, \sum_{k=5}^\infty \,\frac{k^2}{3^k}$$
But the left hand side is a geometric series with common ratio $2/3$, which we know converges. Therefore the right hand side (your sum) must converge. 

Now that we know that it converges, suppose you'd like to find the exact value of the sum. 
Let $S$ be the value to which your sum converges.  
\begin{align}
S&=\sum_{k=1}^\infty \,\frac{k^2}{3^k}\\\\
S&=0 + \frac{1}{3} +\frac{4}{9} + \frac{9}{27}+\frac{16}{81}+\frac{25}{243}+ \cdots\\\\
3S&= 1 + \frac{4}{3} +\frac{9}{9} + \frac{16}{27}+\frac{25}{81}+\frac{36}{243}+ \cdots
\end{align}
Now subtract the second-to-last equation above from the last equation. Let $R = 2S$ be this new series. 
\begin{align}
3S - S &= 1 + \frac{3}{3} + \frac{5}{9} + \frac{7}{27} + \frac{9}{81} + \frac{11}{243} + \cdots \qquad \\\\
R &= 1 + \frac{3}{3} + \frac{5}{9} + \frac{7}{27} + \frac{9}{81} + \frac{11}{243} + \cdots\\\\
3R &= 6 + \frac{5}{3} + \frac{7}{9} + \frac{9}{27} + \frac{11}{81} + \frac{13}{243} + \cdots
\end{align}
Now subtract the second-to-last equation above from the last equation. Let $T = 2R = 4S$ be this new series.
\begin{align}
3R - R &= 5 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} + \frac{2}{243} + \cdots \qquad \\\\
T &= 3 + 2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} + \frac{2}{243} + \cdots\\\\
T &= 3  + 2\left(\sum_{k=0}^\infty \frac{1}{3^k}\right)
\end{align}
We've reached a geometric series with first term $1$ and common ratio $1/3$. This converges to $3/2$, so we are able to go back up the rabbit hole to find the value of $S$, at long last:
\begin{align}
4S &= 2R = T = 3 + 2(3/2) = 6 \qquad\qquad \\\\
S&=\boxed{\frac{3}{2}}
\end{align}

It's worth noting that series like the intermediate $R$ encountered here are called arithmetico-geometric series because the terms' numerators are in arithmetic progression, while the denominators are in geometric progression. 
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