# Convergence and sum of series with exponents

So the question is how can I see if this series :

$$\displaystyle\sum_{n=1}^{\infty} \frac{1}{(4-(-1)^n)^n}$$

converges and find its sum.

So I would probably need to use the Leibnitz criterion for alternating series, but I'm not 100% if this series is an alternating one. Any help would be appreciated.

$$\frac{1}{(4-(-1)^n)^n}\leq \frac{1}{3^n}$$

And comparison test.

It's a positive series, clearly not an alternating one!

Consider the terms for even $n$ and odd $n$ separately, then you get two geometric series.

Your series is not an alternating one. You can rewrite it as $\sum_{n=1}^{\infty} (\frac{1}{5})^{2n-1} + (\frac{1}{3})^{2n}$ which should help you prove its convergence and find the limit

• Can you show the steps as to how you rewrote the series? Thanks! Jun 28, 2016 at 14:44
• Like the others said, separate odd and even terms. Jun 28, 2016 at 14:49

HINT

Look at the first few terms of the series:

$$\frac{1}{(4-(-1))^n}, \frac{1}{(4-1))^n}, \frac{1}{(4-(-1))^n} \ldots$$

The above is for $n=1, 2, 3$. Do you notice a pattern? Can you see why it is not alternating?

As for convergence, the best (quite probably the simplest) option you have is comparison test.