0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ Jun 6 '17 at 21:08
2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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
2
$\begingroup$

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 $\{b_n\}$).

$\endgroup$
1
$\begingroup$

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: $$=\frac{2}{5} \frac{1}{1-\frac{1}{5}}-\frac{1}{5} \frac{1}{1+\frac{1}{5}} =\frac{1}{2} - \frac{1}{6} =\frac{1}{3}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.