# how does $\sum_{n=0}^{\infty} (-1)^n \frac{1}{1+n}$ diverge?

I thought that to prove an alternating series two tests needed to be proven

$$a_n \ge a_{n+1}$$

which is true and

$$\lim_{n\to\infty} b_n = 0 \ \ \ \ \ \ \text{which} \ \ \ \ \ \ \lim_{n\to\infty}\frac{1}{1+n}=0$$

yet sources (wolfram alpha) indicate that it does not converge

• Sources are wrong. What sources? Dec 11, 2013 at 8:38
• Which sources? The series does converge (not absolutely, though).
– Did
Dec 11, 2013 at 8:38
• can you explicitly mention the sources? please Dec 11, 2013 at 8:39
• As the previous somments ask, please provide us explicitly the sources. What they say is totally wrong. Dec 11, 2013 at 8:42
• why an upvote for this? :P
– user87543
Dec 11, 2013 at 9:06

The series is convergent, moreover we can obtain that: \begin{align} &\phantom{={}}\sum_{n=0}^{\infty} (-1)^n \frac{1}{1+n} \\ &=1-\frac12+\frac13-\frac14+\cdots \\ &=\ln2 \end{align} cause $$\ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \frac{1}{1+n}x^n$$ and the original series is $\ln(1+1)$.
In fact it converges. Let $b_n=(-1)^n$ and $a_n=\frac {1}{n+1}$. Then $a_n>0$ is decreasing that goes to $0$.Also $b_n$ has bounded partial sums because $\sum_{k=0}^{n} b_n\leq 1$.So using Dirichlet's Proposition we have that it converges. See here http://en.wikipedia.org/wiki/Dirichlet's_test