# Determine the Convergence of the series $\sum_{k=1}^\infty \frac{2+(-1)^k}{5^k}$ [closed]

So I've been searching for a way to determine whether or not the following series:

$$\sum_{k=1}^\infty \frac{2+(-1)^k}{5^k}$$

converges or not.

I've used the integral test, the comparison test, the limit comparison test, the ratio test and the root test and I didn't manage to accomplish anything.

## closed as off-topic by Zain Patel, Jack D'Aurizio, Shailesh, Namaste, LeucippusJun 7 '17 at 0:37

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• Can you see that $2\sum_{k\geq 1}\left(\frac{1}{5}\right)^k + \sum_{k\geq 1}\left(-\frac{1}{5}\right)^k$ is obviously convergent? – Jack D'Aurizio Jun 6 '17 at 21:08

I see two geometric series: $$\sum_{k=1}^\infty \frac{2+(-1)^k}{5^k} = \frac{2}{5} \sum_{k=0}^\infty (\frac{1}{5})^k-\frac{1}{5} \sum_{k=0}^\infty (\frac{-1}{5})^k$$ Both geometric series are convergent and their combined value is: $$=\frac{2}{5} \frac{1}{1-\frac{1}{5}}-\frac{1}{5} \frac{1}{1+\frac{1}{5}} =\frac{1}{2} - \frac{1}{6} =\frac{1}{3}$$
Hint: $-1 \leq (-1)^k \leq 1$.
Recall that the geometric series, $$\sum_{n=1}^\infty ax^n=\frac{ax}{1-x}$$ converges if and only if $|x|<1$, where $a\in\Bbb R$. So, compare the series you have with the geometric series where $x=\frac{1}{5}$.
Addendum: in fact, using the above formula for the geometric series, one can compute the explicit value of this series by splitting the given series into two geometric series. Note that $$\sum_{n=1}^\infty (a_n+b_n)=\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty b_n$$ provided the individual series converge (i.e. the limits of partial sums exist individually for $\{a_n\}$ and $\{b_n\}$).