Question: $$\sum\_{n=1}^{\infty}({\cos{\frac{a}{n^3}})^{n^3}}$$where a is constant. I have done the next:$$(\cos{\frac{a}{n^3}})^{n^3}\sim{e^{-0.5a^2n}}$$And what is the next step?

Answer: 
**Hint**: The necessary condition for convergence of a series is that after large *n*, the terms should not affect the sum significantly! That is, if for large *n*, the terms should approach zero. Mathematically, we may say that if $\sum\limits\_{n=0}^{\infty} a\_n$ is a series, then it will not converge if $\lim\limits\_{n \rightarrow \infty} a\_n \neq 0$.

Use this for your sequence, it becomes:-
$$\lim\limits\_{n \rightarrow \infty} a\_n = \lim\limits\_{n \rightarrow \infty} \left( \cos \dfrac{a}{n^3} \right)^{n^3} = 1 \neq 0$$.
Hence, the given series is divergent!
$\sum\_{n=1}^{\infty}({\cos{\frac{a}{n^3}})^{n^3}}
$

Since
$\cos(x)
\approx 1-\frac{x^2}{2}
$
for small $x$,
$\cos{\frac{a}{n^3}}
\approx 1-\frac{a^2}{2n^6}
$
so

$\begin{array}\\
(\cos{\frac{a}{n^3}})^{n^3}
&\approx (1-\frac{a^2}{2n^6})^{n^3}\\
&= ((1-\frac{a^2}{2n^6})^{n^6})^{1/n^3}\\
&\approx (e^{-a^2/2})^{1/n^3}\\
&\approx e^{-a^2/2n^3}\\
&\approx 1-\frac{a^2}{2n^3}\\
&\to 1\\
\end{array}
$

so the sum diverges.

Note that if the sum was
$\sum\_{n=1}^{\infty}(1-({\cos{\frac{a}{n^3}})^{n^3}})
$
then it would converge,
since the terms would be
of order $\frac1{n^3}$.