# How to determine the convergence of $\sum_{n\geq 2}{} \frac{(-1)^{n}}{(-1)^n+n}$? [duplicate]

This is most likely very easy to show, but with the load of midterms I had, my brain just declines to work properly. How do I determine the convergence of $$\sum_{n\geq 2}{} \frac{(-1)^{n}}{(-1)^n+n}$$?

## marked as duplicate by Matthew Conroy, Davide Giraudo, Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 5 '16 at 23:44

• One may prove that $\sum_{n\geq 2}{} \frac{(-1)^{n}}{(-1)^n+n}=-1+\ln 2$. – Olivier Oloa Dec 5 '16 at 20:59
• – Martin Sleziak Dec 5 '16 at 23:33

HINT:

\begin{align} \sum_{n=2}^{2N}\frac{(-1)^n}{(-1)^n+n}&=\sum_{n=1}^{N}\left(\frac{1}{2n+1}-\frac{1}{2n}\right)+\frac{1}{2N}\\\\ &=\frac1{2N}-\sum_{n=1}^N \frac{1}{2n(2n+1)} \end{align}

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

First, since $\left|\sum_{n=1}^N\frac{1}{2n(2n+1)}\right|\le \frac{1}{4}\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{24}$, we see that the series of interest converges. In fact, we can evaluate the series in closed form. Proceeding, we write \begin{align}\sum_{n=2}^{2N}\frac{(-1)^n}{(-1)^n+n}&=\sum_{n=1}^N\left(\frac{1}{2n+1}-\frac{1}{2n}\right)+\frac1{2N}\\\\&=\sum_{n=1}^N\left(\frac{1}{2n+1}+\frac{1}{2n}\right)-\sum_{n=1}^N\frac1n+\frac1{2N}\\\\&=-1+\sum_{n=1}^{2N+1}\frac1n-\sum_{n=1}^N\frac1n+\frac1{2N}\\\\&=-1+\sum_{n=1}^{N+1}\frac{1}{n+N}+\frac1{2N}\\\\&=-1+\frac1N\sum_{n=1}^{N+1}\frac{1}{1+n/N}+\frac{1}{2N} \tag {A1}\\\\&\to -1+\int_0^1 \frac{1}{1+x}\,dx\,\,\text{as}\,\,N\to \infty \tag{A2}\\\\&=-1+\log(2)\end{align}where we used only elementary arithmetic to take us to $(A1)$ and recognized the sum in $(A1)$ as a Riemann sum to arrive at $(A2)$. An alternative way forward to evaluating the series is to write $$\sum_{n=1}^N\left(\frac{1}{2n+1}-\frac{1}{2n}\right)+\frac1{2N}=-1+\sum_{n=1}^{2N+1}\frac{(-1)^{n-1}}{n}+\frac{1}{2N}$$Then, recalling that $\log(1+x)$ has Taylor series representation $\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}$ for $-1<x\le 1$, we see that $$\sum_{n=2}^\infty\frac{(-1)^n}{(-1)^n+n}=-1+\log(2)$$as expected!

One may write, by a Taylor series expansion, as $n \to \infty$, $$\frac{(-1)^{n}}{(-1)^n+n}=\frac{(-1)^{n}}{n}\cdot\frac{1}{1+\frac{(-1)^{n}}{n}}=\frac{(-1)^{n}}{n}\left(1-\frac{(-1)^{n}}{n}+O\left(\frac1{n^2}\right) \right)$$ that is, as $n \to \infty$, $$\frac{(-1)^{n}}{(-1)^n+n}=\frac{(-1)^{n}}{n}-\frac{1}{n^2}+O\left(\frac1{n^3}\right)$$ giving that the initial series is convergent being the sum of three convergent series.

• As usual, asymptotic computation yields a straightforward solution. – Gabriel Romon Dec 5 '16 at 21:09
• @LeGrandDODOM So does simple grouping, which is suitable to a pre-calculus student. – Mark Viola Dec 5 '16 at 21:45

It definitely converges; one temptation (which doesn't obviously work though) is to use the alternating series test. However, the terms that show up run as $$1/3, -1/2, 1/5, -1/4, 1/7, -1/6, \ldots$$ which is not a decreasing sequence (upon taking absolute values). However, we can still determine convergence of this by grouping the terms in pairs as $$(1/3 - 1/2) + (1/5 - 1/4) + (1/7 - 1/6) + \cdots$$ which is just the sum $$1/6 + 1/20 + 1/42 + \cdots + \frac{1}{2n(2n+1)} + \cdots$$ which can now be compared to the series $\sum 1/n^2$.

• Using a regroupement only works with additional hypothesis (which are verified here) : the number of terms in each group must be constant, and the general term must have limit $0$. – Nicolas FRANCOIS Dec 5 '16 at 20:56

No it is not. Let's do some asymptotic work on the general term $u_n$ of the series : \begin{align}\frac{(-1)^n}{(-1)^n+n} &= \frac{(-1)^n}{n}\frac{1}{1+\frac{(-1)^n}{n}} \\ &= \frac{(-1)^n}{n}\left(1-\frac{(-1)^n}{n}+o(1/n)\right) \\ &= \frac{(-1)^n}{n} - \frac{1}{n^2}+o(1/n^2)\end{align} So $u_n$ is the sum of an alternate series (the term $\frac{(-1)^n}{n}$), convergent by Leibniz criterion, and of an absolutely convergent series (the group $-\frac{1}{n^2}+o(1/n^2)$), so it's a convergent series.

What you should absolutely not say : $u_n$ is equivalent to $\frac{(-1)^n}{n}$, which is a convergent series, so $\sum u_n$ is convergent. The comparison theorem only works with absolutely convergent series (and in general with constant sign series).

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

$$=\frac{(-1)^n}{n}-\frac{1}{n^2}(1+\epsilon(n))$$

$$=v_n+w_n$$

$\sum v_n$ is convergent as alternate.

$\sum w_n$ is absolutly convergent since $|w_n|\sim \frac{1}{n^2}\;(n\to+\infty)$.

As a sum of two convergent series, $\sum u_n$ converges.

Note that if we group the terms in pairs we get ${(-1)^n \over n+(-1)^n } - {(-1)^n \over n-(-1)^n } = -{ 2 \over n^2 -1}$.

Since $|-{ 2 \over n^2 -1}| \le {4 \over n^2}$ we see that the series converges conditionally.

• Copper HAPPY 100000 – hamam_Abdallah Dec 5 '16 at 21:35
• @AbdallahHammam: Not quite there yet but thanks! :-) – copper.hat Dec 5 '16 at 21:36