Art of Problem Solving Forum

jmerry

Evaluate $\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{1+nk}$ as a function of $k$ .

**Posted:** Fri Dec 17, 2004 7:14 pm

kunny

My idea is $\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{1+nk}=\sum_{n=-\infty}^{\infty} (-1)^n \int^1_0 x^{kn} dx=\int^1_0 \sum_{n=-\infty}^{...$

Especially for $k=2,\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}=\frac{\pi}{4}$ . Libnitz series.By the way do we have another condition for $k$ ?

**Posted:** Fri Dec 17, 2004 7:38 pm

jmerry

That could be useful, but there's a flaw: $\sum_{n=-\infty}^{\infty}x^n$ diverges for all $x$ . I'm being vague about $k$ : the series diverges when $k=\frac1n$ , but that's trivial. It may be easier to assume $k>1$ .

For $k=2$ , you get $\dots-\frac17+\frac15-\frac13+1+1-\frac13+\frac15-\frac17+ \dots=\frac{\pi}{2}$ : the terms for negative $n$ match the positive ones. In general, you can't evaluate the sums for $n>0$ and for $n<0$ independently, although they both converge.

**Posted:** Fri Dec 17, 2004 10:30 pm

kunny

That's true.Thank you, jimrrey.

kunny

**Posted:** Fri Dec 17, 2004 10:51 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

i guess we could use the poisson identity $\sum_{n\in\mathbb{Z}} f(n)=\sum_{k\in\mathbb{Z}} g(k)$ , where $g(k)=\int_{-\infty}^{\infty} f(z)e^{-2\pi ikz}dz$ .

if we take $f(n)=\frac{(-1)^n}{1+xn}=\frac{e^{\pi in}}{1+xn}$ , we get $g(k)=\int_{-\infty}^{\infty} \frac{e^{\pi iz(1-2k)}}{1+xz}dz$ . now, we get the identity $\phi (x)=\sum_{n\in\mathbb{Z}} \frac{(-1)^n}{1+xn}=\sum_{k\in\mathbb{Z}} \int_{-\infty}^{\infty} \frac{e^{\pi iz(1-2k)}}{1+xz...$ . now, assume $x=a+ib$ with $b>0$ . then the integrand has one pole at $z=-\frac{1}{x}$ with imaginary part $>0$ . if $k<1$ , we integrate over the real line and lower complex-halfplane(the value goes to zero there), and there is no pole, thus the integral is 0. if $k>0$ , then we integrate over the real line and the upper complex-halfplane(mathematically positive directed). the pole hat $z=-\frac{1}{x}$ is $\frac{1}{x}e^{\frac{\pi i(2k-1)}{x}}$ ; the value of the integral is thus $\frac{2\pi i}{x}e^{\frac{\pi i(2k-1)}{x}}$ . summing this for $k>0$ gives $\frac{2\pi i}{x}e^{\frac{\pi i}{x}}\left(e^{\frac{2\pi i}{x}}-1\right)^{-1}$ . by analytic continuation, this is the value for all complex $x$ .

Peter

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

btw, this simplifies to $\frac{\frac{\pi}{x}}{\sin(\frac{\pi}{x})}$ . the series converges iff $\frac{1}{x}\not\in\mathbb{Z}$ .

**Posted:** Sat Dec 18, 2004 6:08 am

Myth

BTW, it is the same as $\int_0^{+\infty}\frac{dt}{1+t^x}$ for $x\geq 1$ , $x\in \mathbb{R}$ .
Is there some relation?

**Posted:** Sat Dec 18, 2004 6:12 am

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Myth is out of here

kunny

Probably Complex Integration will do well. We offer to make Kent Merryfield's entrance.

kunny

**Posted:** Sat Dec 18, 2004 7:34 am

Kent Merryfield

You don't need me to make an entrance. I would have said something about the Poisson summation formula, but Peter Scholze took care of that. Yes, Myth, it was inspired by that problem, but jmerry can explain it himself without needing my help.

Myth

Rrrr... Myth

**Posted:** Sat Dec 18, 2004 8:19 am

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Myth is out of here

jmerry

Nice solution, Peter. I didn't have that method, although I do have one that uses analytic continuation to get from the real line to everywhere else.

Now, on the problem itself:
First, the sum does come from that integral:

Assuming $\Re(k)>1$
$\int_0^{\infty}\frac1{1+x^k}dx= \int_0^1(1-x^k+x^{2k}-x^{3k}+\dots)dx + \int_1^{\infty}(x^{-k}-x^{-2k}+x^{-3k}-x^{-4k}+\dots)...$ ., using the geometric series formula to expand $\frac1{1+x^k}$ in two different ways.
The first integral is $\sum_{n=0}^{\infty}\frac{(-1)^n}{1+nk}$ and the second is $\sum_{n=1}^{\infty}\frac{(-1)^n}{1-nk}$ , so the whole integral
$\int_0^{\infty}\frac1{1+x^k}dx= \sum_{n=-\infty}^{\infty}\frac{(-1)^n}{1+nk}$

My initial method was to evaluate this integral in another way (assuming that $k$ is real and $k>1$ ): take the line integral around the wedge with spokes on the real line and at an angle of $\frac{2\pi}{k}$ above, completed by a distant circle back. Since $k>1$ , the integral over the distant circle goes to zero and we have $(1-e^{2\pi i/k})\int_0^{\infty}\frac1{1+x^k}dx=2\pi i\cdot$ (residue at the pole).

The pole is at $e^{\pi i/k}$ , and it has residue $-\frac{e^{\pi i/k}}{k}$ , so $\int_0^{\infty}\frac1{1+x^k}dx=\frac{-2\pi i e^{\pi i/k}}{k(1-e^{2\pi i/k})}=\frac{\pi/k}{\sin(\pi/k)}$

Once we have this, the general version follows by analytic continuation.

Alternately, we can work directly with the sum and never invoke integrals at all.
Let $g(z)=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{z+n}$ . If $f(z)=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{1+nz}$ , $g(z)=\frac{f(\frac1z)}{z}$ .
We can easily see that $g$ is analytic; the alternate form $g(z)=\sum_{n=-\infty}^{\infty}\frac1{(z+2n)(z+2n+1)}$ converges absolutely. Obviously $g$ has simple poles at every integer, with residue $(-1)^n$ . Also, $g(z+1)=-g(z)$ , and $g(z)$ tends to 0 as the imaginary part of $z$ tends to $\pm \infty$ (use the alternate form).
A function that shares these properties is $\frac{\pi}{\sin(\pi z)}$ ; we want to show that it is the same. Consider the function $g(z)- \frac{\pi}{\sin(\pi z)}$ on the vertical strip $|\Im(z)|\le \frac12$ . Matching the Laurent series at the pole, the remaining terms of the difference all have nonnegative exponents, so this singularity is removable; remove it. The function must then be bounded on the whole strip, since it becomes small far enough away. By periodicity, this makes the difference bounded on the whole plane. A bounded entire function is constant, and this one goes to zero somewhere, so it must be zero.

**Posted:** Sat Dec 18, 2004 12:07 pm

Myth

Integral $\int_0^{\inty}\frac{dx}{1+x^k}$ is calculated somewhere on forum by me Wink

That is why I know it by sight.

**Posted:** Sat Dec 18, 2004 12:13 pm

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Myth is out of here