Is $\sum_{n=1}^{\infty}(-1)^n\frac {e^n}{n}$ convergent? Is the below series convergent? How should I find out so?
$$\sum_{n=1}^{\infty}(-1)^n\frac {e^n}{n}$$
I tried to use Leibniz test but I failed finding the right answer.
 A: Hint:$$\lim_{n\to +\infty} \frac {e^n}{n}\neq 0$$
A: Is the limit of the absolute value of the term equal to zero?
A: Your series $$\sum_{n = 1}^{\infty} (-1)^n\frac {e^n}{n}$$ diverges because of the use of the series ratio test (which someone suggested you use). Remember, that test says if $\exists \ \text {an} \ N \ \text {so that} \ \forall \ n \geq N, a_n \neq 0, \text {and}$ $$L = \lim_{n \to \infty} \lvert \frac {a_{n + 1}}{a_n} \rvert$$ the series converges if $L < 1$. If $L > 1$, then the series diverges. If $L = 1$, the series is inconclusive. To prove this, let's work the limit out: $$\lvert \frac {a_{n + 1}}{a_n} \rvert = \left\lvert \frac{(-1)^{(n + 1)\frac{e^{(n + 1)}}{n + 1}}}{(-1)^n\frac {e^n}{n}}\right\rvert$$ Which simplifies to: $$\left\lvert \frac{en}{n + 1} \right\rvert$$ When we plug that into our limit, we get: $$\lim_{n \to \infty}\left(\left\lvert \frac{en}{n + 1}\right\rvert\right) = e = 2.718281828459045...$$ Since $L = e$, and $e > 1$, that means that $L > 1$. So, the series diverges.
