Infinite Series

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May 22nd 2008, 05:08 AM
Dystopia

Infinite Series

A few infinite series that I thought would be fun.

Find closed form expressions for:

i) $\sum_{r=0}^{\infty} (-1)^{r} \left(\frac{1}{4r+1} + \frac{1}{4r+3} \right)$

ii) $\sum_{r=0}^{\infty} (-1)^{r} \frac{(4r)!}{2^{4r}(2r)!(2r+1)!}$

iii) $\sum_{n=0}^{\infty} \frac{1}{2n+1} \left( \sum_{r=n+1}^{\infty} \frac{(-1)^{r-1}}{r} \right)$

A couple of them are based on other series I've seen, whereas the other I just happened to stumble across when contemplating something else. Hopefully they aren't too common and haven't been posted before.

Obviously some calculus is going to be required.
May 22nd 2008, 06:40 AM
Krizalid

Quote:

Originally Posted by Dystopia

A few infinite series that I thought would be fun.

Find closed form expressions for:

i) $\sum_{r=0}^{\infty} (-1)^{r} \left(\frac{1}{4r+1} + \frac{1}{4r+3} \right)$

Sum equals $\sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left( \int_{0}^{1}{x^{4r}\,dx}+\int_{0}^{1}{x^{4r+2}\,dx } \right)}.$

From here $\sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left\{ \int_{0}^{1}{x^{4r}\left( 1+x^{2} \right)\,dx} \right\}},$ hence $\int_{0}^{1}{\left\{ \left( 1+x^{2} \right)\sum\limits_{r\,=\,0}^{\infty }{\left( -x^{4} \right)^{r}} \right\}\,dx}$ and we happily get that the original sum equals $\int_{0}^{1}{\frac{1+x^{2}}{1+x^{4}}\,dx}.$

There's your closed form.
May 22nd 2008, 07:15 AM
Dystopia

Quote:

Originally Posted by Krizalid

Sum equals $\sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left( \int_{0}^{1}{x^{4r}\,dx}+\int_{0}^{1}{x^{4r+2}\,dx } \right)}.$

From here $\sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left\{ \int_{0}^{1}{x^{4r}\left( 1+x^{2} \right)\,dx} \right\}},$ hence $\int_{0}^{1}{\left\{ \left( 1+x^{2} \right)\sum\limits_{r\,=\,0}^{\infty }{\left( -x^{4} \right)^{r}} \right\}\,dx}$ and we happily get that the original sum equals $\int_{0}^{1}{\frac{1+x^{2}}{1+x^{4}}\,dx}.$

There's your closed form.

OK, how about an actual value. I don't blame you if you don't feel like expanding that using partial fractions, but there are nicer ways.
May 22nd 2008, 11:42 AM
PaulRS

Quote:

Originally Posted by Dystopia

OK, how about an actual value. I don't blame you if you don't feel like expanding that using partial fractions, but there are nicer ways.

$ \int_0^1 {\frac{{1 + x^2 }} {{1 + x^4 }}dx} \underbrace = _{u = \tfrac{1} {x}} - \int_{ + \infty }^1 {\left( {\tfrac{1} {{u^2 }}} \right)\frac{{1 + \tfrac{1} {{u^2 }}}} {{1 + \tfrac{1} {{u^4 }}}}du} = \int_1^{ + \infty } {\frac{{u^2 + 1}} {{u^4 + 1}}du} $

Thus: $ I = \int_0^1 {\frac{{1 + x^2 }} {{1 + x^4 }}dx} \Rightarrow 2 \cdot I = \int_0^{ + \infty } {\frac{{1 + x^2 }} {{1 + x^4 }}dx} $

By the linearity of the Integral: $ 2 \cdot I = \int_0^{ + \infty } {\frac{{dx}} {{1 + x^4 }}} + \int_0^{ + \infty } {\frac{{x^2 }} {{1 + x^4 }}dx} $

Now note that: $ \int_0^{ + \infty } {\frac{{x^2 }} {{1 + x^4 }}dx} \underbrace = _{u = \tfrac{1} {x}}\int_0^{ + \infty } {\frac{{du}} {{1 + u^4 }}} $

Thus: $ I = \int_0^{ + \infty } {\frac{{dx}} {{1 + x^4 }}} = \tfrac{\pi } {{2 \cdot \sqrt 2 }} $

Where we used the formula: $ \int_0^{ + \infty } {\frac{{dx}} {{1 + x^n }}} = \tfrac{{\tfrac{\pi } {n}}} {{\sin \left( {\tfrac{\pi } {n}} \right)}} $ with $ n \in \mathbb{N},n \geqslant 2 $
May 22nd 2008, 05:24 PM
NonCommAlg

Quote:

Originally Posted by Dystopia

ii) $\sum_{r=0}^{\infty} (-1)^{r} \frac{(4r)!}{2^{4r}(2r)!(2r+1)!}$

using this simple fact that $\frac{1}{2^{4r}}\binom{4r}{2r}=\frac{2}{\pi}\int_0 ^{\frac{\pi}{2}} \sin^{4r} t \ dt,$ we will get:

$I=\sum_{r=0}^{\infty}(-1)^r \frac{(4r)!}{2^{4r}(2r)!(2r+1)!}=\frac{2}{\pi}\sum _{r=0}^{\infty} \int _0^1 \int_0^{\frac{\pi}{2}} (-x^2)^r\sin^{4r}t \ dt \ dx$

$= \frac{2}{\pi} \int_0^1 \int_0^{\frac{\pi}{2}} \sum_{r=0}^{\infty} (-x^2 \sin^4 t)^r \ dt \ dx = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \int_0^1 \frac{1}{1 + x^2 \sin^4 t} \ dx \ dt$

$= \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \frac{\tan^{-1}(\sin^2 t)}{\sin^2 t} \ dt=\frac{4}{\pi} \int_0^{\frac{\pi}{2}} \frac{\cos^2 t}{1 + \sin^4 t} \ dt.$ now let $\tan t = u.$ then:

$I=\frac{4}{\pi}\int_0^{\infty} \frac{1}{2u^4 + 2u^2 + 1} \ du=\frac{2}{\pi}\int_0^{\infty} \frac{\sqrt{2}+u^{-2}}{(\sqrt{2}u - u^{-1})^2 + 2\sqrt{2} + 2} \ du \ -$

$\frac{2}{\pi} \int_0^{\infty} \frac{\sqrt{2}-u^{-2}}{(\sqrt{2} u + u^{-1})^2 -(2\sqrt{2}-2)} \ du=\sqrt{2\sqrt{2}-2}. \ \ \ \square$

Quote:

iii) $\sum_{n=0}^{\infty} \frac{1}{2n+1} \left( \sum_{r=n+1}^{\infty} \frac{(-1)^{r-1}}{r} \right)$

$\sum_{n=0}^{\infty} \frac{1}{2n+1} \left(\sum_{r=n+1}^{\infty} \frac{(-1)^{r-1}}{r} \right)=\sum_{n=0}^{\infty} \frac{1}{2n+1} \sum_{r=n+1}^{\infty}\int_0^1 (-x)^{r-1} \ dx$

$=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \int_0^1 \frac{x^n}{x+1} \ dx=2 \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_0^1 \frac{t^{2n+1}}{t^2 + 1} \ dt$

$=2 \int_0^1 \frac{1}{t^2+1} \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n+1}}{2n+1} \ dt=2 \int_0^1 \frac{\tan^{-1} t}{t^2 +1} \ dt = \frac{\pi^2}{16}. \ \ \ \square$
May 22nd 2008, 07:07 PM
ThePerfectHacker

Quote:

Originally Posted by Dystopia

i) $\sum_{r=0}^{\infty} (-1)^{r} \left(\frac{1}{4r+1} + \frac{1}{4r+3} \right)$

Note that,

$\sum_{n=-\infty}^{-1} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right) = \sum_{n=-\infty}^{-1}(-1)^{(1-n)-1}\left( \frac{-1}{4(1-n)+3}+\frac{-1}{4(1-n)+1} \right)$ $= \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right)$
Therefore,

$\sum_{n=-\infty}^{\infty}(-1)^n \left( \frac1{4n+1}+\frac1{4n+3}\right) = 2\sum_{n=0}^{\infty}(-1)^n\left( \frac1{4n+1}+\frac1{4n+3}\right)$ .

Now define $f(z) = \frac{1}{4z+1}+\frac{1}{4z+3}$ this function has poles at $\xi = -1/4$ and $\mu = -3/4$ .

If $F(z) = \pi f(z) \csc \pi z$ then by complex analysis methods we have:

$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = - \mbox{Res}(F,\xi) - \mbox{Res}(F,\mu) = \frac{\pi}{2\sqrt{2}}+\frac{\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$ .

Thus, $\sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right) = \frac{1}{2}\cdot \frac{\pi}{\sqrt{2}}$ .
May 23rd 2008, 02:01 PM
Dystopia

Those answers are correct. The methods used are interesting. Here is how I did them:

i) If $\sin\theta > 0$ , then $\sum_{r=0}^{\infty} \frac{\sin(2r+1)\theta}{2r+1} = \frac{\pi}{4}$ .

Put $\theta = \frac{\pi}{4}$

ii) By De Moivre's theorem, $\cos\frac{x}{2} = \Re (\cos x + i\sin x)^{-\frac{1}{2}}$

$\Rightarrow \cos\frac{x}{2}\sqrt{\cos x} = \sum_{r=0}^{\infty} \frac{(-1)^{r} (4r)!}{2^{4r}(2r)!^{2}} \tan^{2r}x \;\;\;\; (|\tan x| \leq 1)$

Multiply by $\sec^{2}x$ and integrate with limits pi/4, 0.

iii) Same as NCA.