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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

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

2 Answers 2

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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 $\ln(1+1)$.

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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

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