Answer
$$\dfrac{5}{6}$$
Work Step by Step
A geometric series $\Sigma_{k=0}^{\infty} ar^k$ is said to be converges when $|r| \lt 1 $ and diverges when $|r| \geq 1$ and $\Sigma_{k=0}^{\infty} ar^k=\dfrac{a}{1-r}$
The given series is a geometric series as $\Sigma_{k=1}^{\infty} ar^k$ with common ratio $r=-\dfrac{1}{5}$. We see that $|r| \lt 1$, so the series converges.
Also, its sum can be computed as:
$\Sigma_{k=1}^{\infty} (-\dfrac{1}{5})^k=\dfrac{1}{1-(-\frac{1}{5})}\\=\dfrac{5}{6}$
