Proving sequence is Cauchy sequence

Question

I need to examine if this sequence is Cauchy sequence using definition. $$a_n= 1/3 + 2^2/3^2+...+n^2/3^n$$ I start with $m,n \in N, n>m$ $$ \vert(a_n - a_m)\vert = \vert(m+1)^2/3^{(m+1)}+...+n^2/3^n\vert$$ And I don't know what is bigger than this expression and what to compare it to.

Answer 1

An idea to make things simpler: take $\;n,\,\,m=n+p\;,\;\;p\in\Bbb N\;$ , so

$$\left|a_{n+p}-a_n\right|=\frac{(n+1)^2}{3^{n+1}}+\ldots+\frac{(n+p)^2}{3^{n+p}}\le p\frac{n^2}{3^n}$$

Since for any $\;p\in\Bbb N\;$ we have that $\;\lim\limits_{n\to\infty}\;p\,\cfrac {n^2}{3^n}=0\;$ , we get that for any $\;\epsilon>0\;$ and for any $\;p\in\Bbb N\;$ there exists $\;N_{p,\epsilon}\in\Bbb N\;$ s.t.

$$n>N_{p,\epsilon}\implies p\frac{n^2}{3^n}<\epsilon$$

Fill in details and complete the proof.