# Does $\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}}{\sqrt {n+1}}}$ converge?

Does the series $\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}}{\sqrt {n+1}}}$ converge or diverge?

The series can be written as $\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n-1}}{\sqrt {n}}}$, which remind me of the alternating harmonic series, wich is not convergent.

The terms are alternating, and therfor shuld the alternating series test be used. Have also checked the Leibniz cri.

Where do I go from here?

• Alternating harmonic series converges. Mar 26 '14 at 16:19
• Thanks for that. But is it wrong to compare it with the alternating harmonic series?
– Ron
Mar 26 '14 at 16:21
• It is not wrong, but I don't think it is useful in this case. Mar 26 '14 at 16:37
• Does that meen that the series also is absolutly convergent? So by applying the Absolutle convergence test would give th same result??
– Ron
Mar 26 '14 at 19:10
• The series $\sum\frac{(-1)^n}{\sqrt{n+1}}$ is not absolutely convergent, it is only conditionally convergent. The series $\sum\frac1{\sqrt{n+1}}$ diverges since this series is equal to $\sum\frac1{n^{1/3}}$ and this series diverges (it is p-series with $p=1/3\le1$). The alternating sign of the series $\sum\frac{(-1)^n}{\sqrt{n+1}}$ makes it convergent. Similarly, the harmonic series $\sum\frac1n$ diverges, but the alternating harmonic series $\sum\frac{(-1)^{n+1}}n$ converges. Mar 26 '14 at 19:49

By alternating series test, the series converges since $$\biggl|\frac{(-1)^n}{\sqrt{n+1}}\biggr|=\frac1{\sqrt{n+1}}$$ decreases monotonically and goes to $0$ in the limit.
$$\sum _{n=0}^{\infty } \frac{(-1)^n}{\sqrt{n+1}}$$ converges for Leibniz criterium. It is alternating and its general term $$\frac{1}{\sqrt{n+1}}\to 0$$ as $$n\to\infty$$
Its sum is $$\left(1-2^{2/3}\right) \zeta \left(\frac{1}{3}\right)$$