Study convergence of the series $\sum\limits_n\left(\left(3+3^{-n^3}\right)^{n^2}-3^{k(n)}\right)$, where $k(n)=\frac{n^3 +n^63^{n^2}}{1+n^43^{n^2}}$

Question

The question is to study the convergence of the series $\sum x(n)$, where $$x(n)=\left(3+3^{-n^3}\right)^{n^2}-3^{k(n)}\qquad k(n)=\frac{n^3 +n^63^{n^2}}{1+n^43^{n^2}}$$

I tried to put in evidence $3^{k(n)}$ or to transform every $3^x$ into $e^{x\log(3)}$, but with no success.

Answer 1

Tools:

• If $a_n\to0$ and $a_nb_n\to0$, then $(1+a_n)^{b_n}-1\sim a_nb_n$.
• If $c>0$ and $d_n\to0$, then $c^{d_n}-1\sim d_n\log c$.

Note that $$x(n)=3^{n^2}\left(y(n)-3^{\ell(n)}\right)$$ with $$y(n)=\left(1+3^{-n^3-1}\right)^{n^2}$$ and $$\ell(n)=k(n)-n^2=\frac{n^3-n^2}{1+n^43^{n^2}}\sim n^{-1}3^{-n^2}$$ Thus, $y(n)\to1$ and $3^{\ell(n)}\to1$ with $$y(n)-1\sim n^23^{-n^3-1}\qquad3^{\ell(n)}-1\sim\ell(n)\log3\sim n^{-1}3^{-n^2}\log3$$ This implies that $$y(n)-3^{\ell(n)}\sim-n^{-1}3^{-n^2}\log3$$ hence $$x(n)\sim-n^{-1}\log3$$ and the series $\sum x(n)$ diverges.