# Convergence or divergence for infinite series $\sum_{n=1}^{\infty} \frac{(-1)^5(-5)^n}{n5^n}$

Can someone please explain why this diverges or converges and explain which test they used? For some reason my mind isn't working today and I am getting really frustrated.

$\displaystyle\sum_{n=1}^{\infty} \dfrac{((-1)^5)((-5)^n)}{n5^n}$

Take out 1 negative sign from $(-5)^n$, and you will see that the series can be written as $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n}$. This is (conditionally) convergent by alternating series test.

• But why does it converge? Isn't that harmonic series? – Nocturne Mar 19 '15 at 2:51
• harmonic series is the series WITHOUT the $(-1)^n$. You can use alternating series test: The absolute value is a decreasing sequencing to 0. – cgo Mar 19 '15 at 2:53

First of All, $(-1)^5 =-1$ and $\frac{(-5)^n}{n5^n} = (\frac{-5}{5})^n(\frac{1}{n}) = (-1)^n\frac{1}{n}$ but then You multiply that by $(-1)$ remember because $(-1)^5 =-1$ and so you have $-\frac{(-1)^n}{n}$ and so it becomes $-\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n}$ and this converges by the alternating series because$\lim_{x \to \infty}{|\frac{(-1)^n}{n}}|= lim_{x \to \infty}{|\frac{1}{n}}|=0$

Check this it explain the alternating series test http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

$S = \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}$ is convergent as an alternating series.