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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.

Can anyone please help me?

Thank you in advance.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Zain Patel, Jack D'Aurizio, Shailesh, Namaste, Leucippus
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  • $\begingroup$ 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? $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 21:08
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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: $$=\frac{2}{5} \frac{1}{1-\frac{1}{5}}-\frac{1}{5} \frac{1}{1+\frac{1}{5}} =\frac{1}{2} - \frac{1}{6} =\frac{1}{3}$$

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Hint: $-1 \leq (-1)^k \leq 1$.

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  • $\begingroup$ Note that I'd be more inclined to give more detail to my answer if you showed us any work or even your failed attempts. $\endgroup$ – JavaMan Jun 6 '17 at 19:56
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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 $\{b_n\}$).

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