Infinite Series

**Dystopia** · May 22nd 2008, 06:08 AM

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.

**Krizalid** · May 22nd 2008, 07:40 AM

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.

**Dystopia** · May 22nd 2008, 08:15 AM

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.

**PaulRS** · May 22nd 2008, 12:42 PM

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 $

**NonCommAlg** · May 22nd 2008, 06:24 PM

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:

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

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

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

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

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

$=2 \int_0^1 \frac{1}{t^2+1} \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n+1}}{2n+1} \ dt=2 \int_0^1 \frac{\tan^{-1} t}{t^2 +1} \ dt = \frac{\pi^2}{16}. \ \ \ \square$

**ThePerfectHacker** · May 22nd 2008, 08:07 PM

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, $\sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right) = \frac{1}{2}\cdot \frac{\pi}{\sqrt{2}}$ .

**Dystopia** · May 23rd 2008, 03:01 PM

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.

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