# Determine the convergence of $\sum_{n=2}^{\infty} \frac{(-1)^n}{(-1)^n+n}$ [duplicate]

This question already has an answer here:

I've been really stuck on this infinite series.

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

Trying to determine if it converges or diverges.

As far as I've got is seperating it into two partial sums (odds and evens) but even this doesn't help me.

Any ideas?

Thanks

## marked as duplicate by Sil, user99914, José Carlos Santos, Claude Leibovici, NamasteMay 20 '18 at 11:22

• Can you factor $\frac{(-1)^n}{n}$ and proceed to expand the remaining expression to compare it with well-known series? – Raito May 19 '18 at 20:23
• Google Leibniz Series – DonAntonio May 19 '18 at 20:24
• try to see if it converges absolutely – user438666 May 19 '18 at 20:24
• The way I would do it is by applying Abel summation, or summation by parts. The $(-1)^n$ is the part that gets summed, the $\frac{1}{(-1)^n+n}$ the part that gets differentiated. You cannot apply Dirichlet-Abel test directly, because $\frac{1}{(-1)^n+n}$ is not monotone, but that doesn't matter because the series $\sum|a_{n+1}-a_n|$ converges (comparing with $\sum\frac{1}{k^2}$). – user561348 May 19 '18 at 20:49
• – Sil May 19 '18 at 22:32

Looking at the partial sums,

$\begin{array}\\ \sum_{n=2}^{2m+1} \frac{(-1)^n}{(-1)^n+n} &=\sum_{n=1}^{m} \left(\frac{1}{2n+1}-\frac{1}{2n}\right) \qquad 2n \text{ and } 2n+1\\ &=\sum_{n=1}^{m} -\frac{1}{2n(2n+1)}\\ \end{array}$

and this sum converges absolutely by comparison with $\sum \frac1{n^2}$.

To see this in an elementary way, starting at $n=2$ which does not affect convergence,

$\begin{array}\\ \sum_{n=2}^{m} \frac{1}{2n(2n+1)} &\lt \sum_{n=2}^{m} \frac{1}{2n(2n-2)}\\ &= \frac14\sum_{n=2}^{m} \frac{1}{n(n-1)}\\ &= \frac14\sum_{n=2}^{m} (\frac{1}{n-1}-\frac1{n})\\ &= \frac14(1-\frac1{m})\\ &\lt \frac14\\ \end{array}$

so, restoring $=1$,

$\sum_{n=1}^{m} \frac{1}{2n(2n+1)} \lt \frac16+\frac14 =\frac13$.

Note: Wolfy says the sum is $1-\ln(2) \approx 0.306852819440054690582767878541823431924499865639744745879...$

• Yes, same idea! So... – dan_fulea May 19 '18 at 21:23

Note that

$$\sum_{n=2}^m \frac{(-1)^n}{(-1)^n + n} = \sum_{n=2}^m \frac{(-1)^n}{(-1)^n + n}\frac{n - (-1)^n}{n - (-1)^n} \\= \underbrace{\sum_{n=2}^m \frac{(-1)^nn}{n^2 - 1}}_{\text{(1)}} - \underbrace{\sum_{n=2}^m \frac{1}{n^2 -1 }}_{\text{(2)}},$$

where (1) is a convergent alternating series and (2) converges since $\frac{1}{n^2-1} \sim \frac{1}{n^2}$

The key word is Leibnitz. But there is a small but, the sequences of the absolute values for which we build the series is not decreasing to zero. One can take care, and still apply Leibnitz... But ok, this is a special problem with a possible special solution, it is always good to "see the series", then maybe even more as a bonus, how quick/slow it converges/diverges...: \begin{aligned} \sum_{n\ge 2}\frac{(-1)^n}{n+(-1)^n} &= \lim_{N\to\infty} \underbrace{\sum_{2\le n\le N}\frac{(-1)^n}{n+(-1)^n}}_{\text{Notation: }a_N} \\ &\qquad\text{ so let us consider an odd index general term a_{2N+1}, N\ge 1} \\ a_{2N+1} &= \left(\frac 13-\frac 12\right) + \left(\frac 15-\frac 14\right) + \left(\frac 17-\frac 16\right) + \dots + \left(\frac 1{2N+1}-\frac 1{2N}\right) \\ &= -\frac 1{2\cdot 3} -\frac 1{4\cdot 5} -\frac 1{6\cdot 7} -\dots- \frac 1{2N(2N+1)}\ . \end{aligned} It is enough to consider the convergence / divergence of this subsequence, because the subsequence of even indexed terms in $(a_N)$ differs from this one by a zero sequence.

Now $(a_{2N+1})$ has negative terms, and is strictly decreasing. We show it is bounded (below), and thus the convergence. I prefer positive numbers, for instance for typographical reasons, so consider $(-a_{2N+1})$, and let us bound it as follows: \begin{aligned} &\frac 1{2\cdot 3} +\frac 1{4\cdot 5} +\dots+ \frac 1{2N(2N+1)} \\ &\qquad\le \frac 1{1\cdot 3} +\frac 1{3\cdot 5} +\dots+ \frac 1{(2N-1)(2N+1)} \\ &\qquad= \frac 12\left(\frac 11-\frac 13\right) + \frac 12\left(\frac 13-\frac 15\right) + \dots + \frac 12\left(\frac 1{2N-1}-\frac 1{2+N}\right) \\ &\qquad=\frac 12\left(\frac 11-\frac 1{2+N}\right) \\ &\qquad\longrightarrow \frac 12 \ . \end{aligned} So we expect that the given sequence is converging to a real number $\ge -\frac 12$.

Note: It is always important in such cases to see the numerical value, pari/gp gives

? sum(n=2, 2000001, 1.0*(-1)^n/(n+(-1)^n) )
%8 = -0.30685256944024219045776794885429218552
? sum(n=1, 1000000, 1.0/(2*n)/(2*n+1) )
%9 = 0.30685256944024219045776794885429144764

• Yup. Same idea I had, so I upvoted you. – marty cohen May 19 '18 at 21:19
• Also, Wolfy says the sum is 1-log(2) which agrees with your answer. – marty cohen May 19 '18 at 21:22

Let's take a look at the 2 series:

(1) $\sum_{n=2}^{\infty}-\frac{1}{n^2+(-1)^nn}$

(2) $\sum_{n=2}^{\infty}\frac{(-1)^n}{n}$

Series (1) converges absolutely (comparing with $\sum\frac{1}{n^2}$).

Series (2) is the Leibniz Series and converges conditionally.

Using limit laws, the series obtained by summing (1) and (2) converges as well:

$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}+\sum_{n=2}^{\infty}-\frac{1}{n^2+(-1)^nn}=\sum_{n=2}^{\infty}\frac{n(-1)^n+(-1)^{2}}{n}=\sum_{n=2}^{\infty}\frac{(-1)^n}{(-1)^n+n}$

We have

$$\frac{(-1)^n}{(-1)^n+n}=\frac{(-1)^n}{n}\left(1+\frac{(-1)}{n}\right)^{-1}=\frac{(-1)^n}{n}\left(1+O\left(\frac1n\right)\right)=\frac{(-1)^n}{n}+O\left(\frac1{n^2}\right)$$

• The series $\sum \frac{(-1)^n}n$ converges by Leibniz criterion
• the series $\sum O\left(\frac1{n^2}\right)$ is convergent by comparison with Riemann series

hence the given series is also convergent.