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How would one go around proving the convergence of: $ \sum \frac{(-1)^n}{n+(-1)^n} $. I'm fairly certain, that this series converges, but it doesn't do so absolutely $\left( \frac{1}{n+(-1)^n} \approx \frac{1}{n}\right)$. Leibniz criterion of convergence can't be used, because $\left( \frac{1}{n+(-1)^n} \right) $ is not nonincreasing.

Note: I've noticed, that by shuffling terms of the series one would get $\sum (-1)^n a_n$, where $a_n$ is a nonincreasing sequence, but shuffling is not allowed since the series doesn't converge absolutely.

Botond
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Vilda
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5 Answers5

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Note that $$\frac{(-1)^n}{n+(-1)^n}=\begin{cases}\frac{1}{2k+1}&\text{if $n=2k$,}\\ -\frac{1}{2k}&\text{if $n=2k+1$.}\end{cases}$$ Hence $$\sum_{n=2}^{N}\frac{(-1)^n}{n+(-1)^n}= \begin{cases}\sum_{k=2}^{N}\frac{(-1)^{k+1}}{k}+\frac{1}{N}+\frac{1}{N+1}&\text{if $N$ is even,}\\ \sum_{k=2}^{N}\frac{(-1)^{k+1}}{k}&\text{if $N$ is odd.}\end{cases}$$ Therefore the given series is convergent and it is convergent to the same sum of $\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k}$ (convergent by Leibniz): $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n+(-1)^n}=\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k}=-1+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}=-1+\ln(2).$$

Robert Z
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  • The only reason I didn't accept this as an answer is that I like the accepted one only an epsilon better. I like the reasoning behind this one though. – Vilda Dec 03 '17 at 10:29
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    @ViliX64 That's fine. However, by using my approach you may conclude that the series is convergent AND it converges to $-1+\ln(2)$ – Robert Z Dec 03 '17 at 10:33
  • How does it follow that it converges to $ -1 + ln(2) $ ? – Vilda Dec 03 '17 at 10:34
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    We use the Taylor series of $\ln(1+x)$. See https://math.stackexchange.com/questions/878374/taylor-series-of-ln1x?noredirect=1&lq=1 – Robert Z Dec 03 '17 at 10:41
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\begin{align}\sum_{k=2}^n\frac{(-1)^n}{n+(-1)^n}&=\frac13-\frac12+\frac15-\frac14+\frac17-\frac16+\cdots\\&=-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\cdots\end{align}and this series converges, by Leibniz criterion. This kind of shuffling is allowed. By “this kind” what I mean is the replacement of a series $\sum_{n=1}^\infty a_n$ by $\sum_{n=1}^\infty a_{\sigma(n)}$ where $\sigma\colon\mathbb{N}\longrightarrow\mathbb N$ is a bijection such that the set $\{n-\sigma(n)\,|\,n\in\mathbb{N}\}$ is bounded.

José Carlos Santos
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  • I don't like shuffling terms like this though, much less keeping equal signs between lines when the original series converges conditionally. – Simply Beautiful Art Dec 03 '17 at 12:41
  • @SimplyBeautifulArt When I wrote that$$\frac13-\frac12+\frac15-\frac14+\frac17-\frac16+\cdots=-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\cdots,$$I didn't mean that the series are equal; it is obvious that they are not. What I meant was that their sums are equal, as in$$\frac12+\frac16+\frac1{12}+\frac1{20}+\cdots=\frac12+\frac14+\frac18+\frac1{16}+\cdots$$ – José Carlos Santos Dec 03 '17 at 12:47
  • :P Yeah I know what you mean. But it's probably a lot easier to make a small hint at partial sums. – Simply Beautiful Art Dec 03 '17 at 12:55
  • Oh, nvm. You have a nice explanation at the end. Excuse my blindness. – Simply Beautiful Art Dec 03 '17 at 12:56
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Notice that

$$ \frac{(-1)^n}{n+(-1)^n} = \frac{(-1)^n}{n} - \frac{1}{n(n+(-1)^n)}.$$

Now

  • $\sum \frac{(-1)^n}{n}$ converges by alternating series test, and
  • $ \sum \frac{1}{n(n+(-1)^n)}$ converges by comparison with $\sum \frac{1}{n^2} < \infty$.

Therefore their difference also converges as well. One may also identify this trick as the following asymptotic expansion

$$ \frac{(-1)^n}{n+(-1)^n} = \frac{(-1)^n}{n\left(1 + \mathcal{O}(n^{-1}) \right)} = \frac{(-1)^n}{n} + \mathcal{O}(n^{-2}) $$

which is summable.

Sangchul Lee
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$$\frac{(-1)^n}{n+(-1)^n}<\frac{(-1)^n}{2 n}, \text{ for any }n\in\mathbb{N}\land n\geq 2$$ As the second series converges for Leibniz criterion, so does the first

Hope this helps

Raffaele
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$a_n: \dfrac{(-1)^n}{n+(-1)^n}= $

$\dfrac{(-1)^n}{n+(-1)^n} \dfrac{n-(-1)^n}{n-(-1)^n}=$

$\dfrac{(-1)^nn -1}{ n^2 -1} ;$

$b_n:= \dfrac{n}{n^2-1}=$

$ 1/2(\dfrac{1}{n+1} +\dfrac{1}{n-1});$

$c_n := -\dfrac{1}{n^2-1}.$

$a_n = (-1)^n b_n + c_n.$

Use Leibniz criterion for alternating series

$\sum (-1)^nb_n$;

Use comparison test $(1/n^2)$ for

$\sum c_n .$

Hence $ \sum a_n $ convergent.

Peter Szilas
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