I'd like to discuss the following problem,
Given the following series $$\sum\limits_{n=1}^{\infty} (e^{n\cdot cos(\pi n)} - n)$$
discuss the convergence,divergence...
The question is how to approach the general method when I discover there are two subsequences that respectively have two different limits :
Noticing that $n \cdot cos(\pi n) = n \cdot (-1)^{n}$ we can see that if $n$ is even $a_{n} = e^{n} - n$, if odd $a_{n} = \frac{1}{e^{n}} - n $
So we found two subsequences,the first one positive, the second one negative, that respectively go to $+\infty,-\infty$.
Is there a general approach to formalise the fact that $a_{2n} $ is growing to $+\infty$ much faster than $a_{2n+1}$ is decreasing to $-\infty$ and thus the series is diverging to $+\infty$ ?
Edit: I read your question wrongly and thought you were discussing about sequence earlier.
Since $$\lim_{n \to \infty}e^{2n}-(2n)+\frac1{e^{2n-1}}-(2n-1)=+\infty$$
The series go to $+\infty$.