# An Alternating Infinite Series

• Jan 23rd 2011, 04:57 AM
TheCoffeeMachine
An Alternating Infinite Series
Challenge Problem: Find $\displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.$

Moderator approved CB
• Jan 23rd 2011, 07:16 AM
alexmahone
Quote:

Originally Posted by TheCoffeeMachine
Challenge Problem: Find $\displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.$

$\displaystyle 16k^2+16k+3=(4k+2)^2-1=(4k+3)(4k+1)$

$\displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}=\sum_{k\ge 0}\frac{(-1)^k}{(4k+3)(4k+1)}$

$\displaystyle = \sum_{k\ge 0}\frac{(-1)^k}{2}(\frac{1}{4k+1}-\frac{1}{4k+3})$

$\displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3})-\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+...$

$\displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+...)$

I'm afraid I don't know how to proceed.
• Jan 23rd 2011, 10:45 AM
chisigma
Let suppose to have the periodic function defined in $-\pi as...

$f(x)=\left\{\begin{array}{ll}1 ,\,\,|x|< \frac{\pi}{4}\\{}\\0 ,\,\, |x|>\frac{\pi}{4}\end{array}\right.$ (1)

... so that it can be written in Fourier series as...

$\displaystyle f(x)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n}\ \cos nx$ (2)

The $a_{n}$ can be computed in the usual way...

$\displaystyle \frac{a_{0}}{2} = \frac{1}{\pi}\ \int_{0}^{\frac{\pi}{4}} dx = \frac{1}{4}$

$\displaystyle a_{n} = \frac{2}{\pi}\ \int_{0}^{\frac{\pi}{4}} \cos nx\ dx = \frac{2}{\pi\ n}\ \sin n\ \frac{\pi}{4}$ (3)

Now we combine the results and obtain ...

$\displaystyle f(\frac{\pi}{4}) = \frac{1}{4} + \frac{2}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin n\ \frac{\pi}{4}\ \cos\ n\frac{\pi}{4}}{n} = \frac{1}{2} \implies$

$\displaystyle \implies \sum_{n=1}^{\infty} \frac{\sin n\ \frac{\pi}{4}\ \cos\ n\frac{\pi}{4}}{n} = \frac{1}{2}\ (1-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + ...) =$

$\displaystyle = \sum_{n=0}^{\infty} \frac{1}{16\ n^{2} + 16\ n +3} = \frac{\pi}{8}$ (4)

Kind regards

$\chi$ $\sigma$
• Jan 23rd 2011, 03:26 PM
Prove It
Quote:

Originally Posted by alexmahone
$\displaystyle 16k^2+16k+3=(4k+2)^2-1=(4k+3)(4k+1)$

$\displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}=\sum_{k\ge 0}\frac{(-1)^k}{(4k+3)(4k+1)}$

$\displaystyle = \sum_{k\ge 0}\frac{(-1)^k}{2}(\frac{1}{4k+1}-\frac{1}{4k+3})$

$\displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3})-\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+...$

$\displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+...)$

I'm afraid I don't know how to proceed.

I didn't see anything wrong with what you had posted...
• Jan 23rd 2011, 05:36 PM
TheCoffeeMachine
Quote:

Originally Posted by alexmahone
I'm afraid I don't know how to proceed.

I don't think it telescopes to any simple value or a familiar series...
On the right track, though. I did exactly the same up to this step:

$\displaystyle \sum_{k\ge 0}\bigg(\frac{(-1)^k}{2(4k+1)}-\frac{(-1)^k}{2(4k+3)}\bigg).$

Quote:

Originally Posted by chisigma
$\displaystyle \implies \sum_{n=1}^{\infty} \frac{\sin n\ \frac{\pi}{4}\ \cos\ n\frac{\pi}{4}}{n} = \frac{1}{2}\ (1-\frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - \frac{1}{11} - \frac{1}{13} + ...) =$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{16\ n^{2} + 16\ n +3} = \frac{\pi}{8}$ (4)

I think what you've found is the sum of the series $\displaystyle \sum_{k\ge 0}\frac{1}{16k^2+16k+3}$, not $\displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.$

EDIT: I'm wondering whether $\displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}$ can be found in the same way? It would be nice!
• Jan 23rd 2011, 10:45 PM
alexmahone
Quote:

Originally Posted by Prove It
I didn't see anything wrong with what you had posted...

I had found the sum of the series $\displaystyle \sum_{k\ge 0}\frac{1}{16k^2+16k+3}$, not $\displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.$
• Jan 23rd 2011, 11:14 PM
simplependulum
First resolve the summand into partial fractions and we can see that it is equal to ( the result first obtained by alex ) $\frac{1}{2} ( \frac{1}{1} - \frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - ... )$

or

$\frac{1}{2} \left[ \int_0^1 ( 1 - x^2(1+x^2) + x^6(1+x^2) - x^{10}(1+x^2) + ... ) ~dx \right]$

$= \frac{1}{2} \int_0^1 \left( 1 - \frac{x^2(1+x^2)}{1+x^4} \right) ~dx$

$= \frac{1}{2} \int_0^1 \frac{1-x^2}{1+x^4}~dx$

$= - \frac{1}{2} \int_0^1 \frac{1 - 1/x^2}{ ( x+1/x)^2 - 2 } ~dx$

Sub. $x + 1/x = t$ we have

$S = \frac{1}{2} \int_2^{\infty} \frac{dt}{t^2 - 2 } = \frac{1}{4\sqrt{2}} \left[ \ln[\frac{t - \sqrt{2} }{t + \sqrt{2}} ] \right]_2^{\infty} = \frac{1}{2\sqrt{2}} \ln(1+\sqrt{2} )$
• Jan 23rd 2011, 11:37 PM
TheCoffeeMachine
Quote:

Originally Posted by simplependulum
...

Correct! I did it in a similar fashion with minor differences:

Writing the sum as an integral:

\begin{aligned}S & = \sum_{k \ge 0}\frac{(-1)^k}{16k^2+16k+3} \\& = \sum_{k \ge 0}\frac{(-1)^k(4k+3)-(-1)^k(4k+1)}{2(4k+3)(4k+1)} \\& = \sum_{k \ge 0}\frac{(-1)^k(4k+3)-(-1)^k(4k+1)}{2(4k+3)(4k+1)} \\& = \sum_{k \ge 0} \bigg(\frac{(-1)^k}{2(4k+1)} -\frac{(-1)^k}{2(4k+3)}\bigg) \\&= \sum_{k \ge 0} \bigg(\frac{1}{2}\int_0^1(-1)^kx^{4k}~ dx -\frac{1}{2}\int_0^1(-1)^kx^{4k+2} ~ dx \bigg) \\& = \int_0^1 \bigg(\frac{1}{2}\sum_{k \ge 0}(-1)^kx^{4k} -\frac{1}{2}\sum_{k \ge 0}(-1)^kx^{4k+2}\bigg)\;dx \\& = \int_0^1 \bigg(\frac{1}{2+2x^4} -\frac{x^2}{2x+2x^4}\bigg)~ dx = \int_0^1\frac{1-x^2}{2+2x^4}\;{dx} \end{aligned}

From there it was just a matter of finding the value of the integral:

\begin{aligned} S & = \frac{1}{2}\int_0^1\frac{1-x^2}{1+x^4}\;{dx} = \frac{1}{2}\int_0^1\frac{1-x^2}{(x^2+1)^2-2x^2}{\;dx} \\& = \frac{1}{2}\int_0^1\frac{1-x^2}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)} ~ {\;dx}\\& = \frac{1}{4\sqrt{2}}\int_0^1\bigg\{\frac{(x^2+\sqrt {2}x+1)'}{(x^2+\sqrt{2}x+1)}-\frac{(x^2-\sqrt{2}x+1)'}{(x^2-\sqrt{2}x+1)}\bigg\} ~ dx \\& = \frac{1}{4\sqrt{2}} \biggl\| \ln\bigg(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\bigg) \biggl\|_{x = 0}^{x = 1} = \frac{1}{4\sqrt{2}}\ln\bigg(\frac{2+\sqrt{2}}{2-\sqrt{2}}\bigg).\end{aligned}
• Jan 24th 2011, 12:11 AM
TheCoffeeMachine
The sum that Alex and Chisigma calculated earlier can also be found in the same way:

\begin{aligned}S & = \sum_{k \ge 0}\frac{1}{16k^2+16k+3} \\& = \sum_{k \ge 0}\frac{(4k+3)-(4k+1)}{2(4k+3)(4k+1)} \\& = \sum_{k \ge 0}\frac{(4k+3)-(4k+1)}{2(4k+3)(4k+1)} \\& = \sum_{k \ge 0} \bigg(\frac{1}{2(4k+1)} -\frac{1}{2(4k+3)}\bigg) \\&= \sum_{k \ge 0} \bigg(\frac{1}{2}\int_0^1x^{4k}~ dx -\frac{1}{2}\int_0^1 x^{4k+2} ~ dx \bigg) \\& = \int_0^1 \bigg(\frac{1}{2}\sum_{k \ge 0}x^{4k} -\frac{1}{2}\sum_{k \ge 0}x^{4k+2}\bigg)\;dx \\& = \int_0^1 \bigg(\frac{1}{2-2x^4} -\frac{x^2}{2x-2x^4}\bigg)~ dx = \int_0^1\frac{1-x^2}{2-2x^4}\;{dx} \end{aligned}

And this integral is rather much simpler than the other one:

\begin{aligned}& S = \int_0^1\frac{1-x^2}{2-2x^4}\;{dx} = \frac{1}{2}\int_0^1 \frac{1-x^2}{(1-x^2)(1+x^2)}\;{dx} \\& = \frac{1}{2}\int_0^{1}\frac{1}{1+x^2}\;{dx} = \frac{1}{2}\biggl\| \arctan{x}\biggl\| _{x=0}^{x=1} = \frac{1}{2}\left(\frac{\pi}{4}\right) = \frac{\pi}{8}.\end{aligned}
• Jan 26th 2011, 05:04 AM
TheCoffeeMachine
It turns out that $\displaystyle \sum_{k \ge 0}\frac{1}{16k^2+16k+3}$ is also a special case of the more general result:

$\displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\ right)}{\sqrt{4\alpha+1}}$

which is very handy and punishes lots of sums, but I don't seem to be able to prove it!
• Jan 26th 2011, 12:45 PM
Krizalid
an alternate computation could be the following:

\begin{aligned}
\sum\limits_{k=0}^{\infty }{\frac{{{(-1)}^{k}}}{(4k+3)(4k+1)}}&=\int_{0}^{1}{\int_{0}^{1 }{{{y}^{2}}\left( \sum\limits_{k=0}^{\infty }{{{\left( -{{x}^{4}}{{y}^{4}} \right)}^{k}}} \right)\,dx}\,dy} \\
& =\int_{0}^{1}{\int_{0}^{1}{\frac{{{y}^{2}}}{1+{{x} ^{4}}{{y}^{4}}}\,dx}\,dy} \\
& =\int_{0}^{1}{\int_{0}^{y}{\frac{y}{1+{{t}^{4}}}\, dt}\,dy} \\
& =\int_{0}^{1}{\int_{t}^{1}{\frac{y}{1+{{t}^{4}}}\, dy}\,dt} \\
& =\frac{1}{2}\int_{0}^{1}{\frac{1-{{t}^{2}}}{1+{{t}^{4}}}\,dt}, \\
\end{aligned}

the last integral is routine, and yields the expected result.