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I want to check, whether $\sum\limits_{n=1}^{\infty}\frac{1+(-1)^n}{n}$ converges or diverges.

Leibniz's test failed, and ratio test just made it even more complicated, so i tried to use the comparison test, but i can't find a suitable series so that $\lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n}$ exists..

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    $\begingroup$ Write down the first few terms, six or so. You will spot a pattern. Prove it and conclude. $\endgroup$ – Daniel Fischer Dec 25 '13 at 21:34
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To prove: $\sum_{n\geq 1} \frac{1+(-1)^n}{n}$ diverges.

Proof: \begin{align*} \sum _{n\geq 1} \frac{1+(-1)^n}{n} &= \sum _{k\geq 1} \frac{1+(-1)^{2k}}{2k} + \sum _{k\geq 1} \frac{1+(-1)^{2k-1}}{2k-1} \\ &= \sum _{k\geq 1} \frac{2}{2k} + \sum _{k\geq 1} \frac{0}{2k-1} \\\ &= \sum _{k\geq 1} \frac{1}{k} \end{align*} Because $\sum _{k\geq 1} \frac{1}{k}$ diverges, $\sum_{n\geq 1} \frac{1+(-1)^n}{n}$ diverges as well. \qed

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Hint: The partial sums are $$S_N=H_{\lfloor N/2 \rfloor} $$ where $H_k$ is the sequence of harmonic numbers, i.e. the partial sums of the (divergent) harmonic series $\sum_{n=1}^\infty 1/n$.

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Since the summation is just an addition of the sequence $a_n=\frac{1+(-1)^n}{n}$, let's examine the sequence $a_n$.

For $n$ odd, we have $a_{\text{odd}}=\frac{1+-1}{n_{odd}}=0$. So the only terms that remain are the even terms which are $a_{even}=\frac{1+1}{n_{even}}=\frac{2}{n_{even}}$. Since $n$ is even, here it is of the form $n=2k$ for $k \in \mathbb{Z}_+$. So the terms of our series which are nonzero are $$ a_{even}=\frac{2}{n_{even}}=\frac{2}{2k}=\frac{1}{k} $$ If we add all these we get $$ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots $$ which is the harmonic series, which clearly diverges.

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It clearly diverges

The sum is $$ 2/2 + 2/4 + 2/6 + 2/8 \cdots = 1+1/2+1/3 \cdots $$

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  • $\begingroup$ But how can i show this formally correct? $\endgroup$ – fear.xD Dec 25 '13 at 21:42
  • $\begingroup$ induction: prove that each term will be of a certain form and simplifies to another form implying that this sum is equivalent to the harmonic sum $\endgroup$ – frogeyedpeas Dec 25 '13 at 21:53

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