Sum to infinity of series

**arze** · July 29th 2010, 06:33 PM

Find the sum of the following series:
$\frac{2}{1}(\frac{1}{3})+\frac{2}{3}(\frac{1}{3})^ 3+\frac{2}{5}(\frac{1}{3})^5+...+\frac{2}{2r-1}(\frac{1}{3})^{2r-1}+...$
I don't know where to begin in this, can't recognize any of the known series.

$1-\frac{2}{4}+\frac{2.4}{4.8}-\frac{2.4.6}{4.8.12}-...$
My attempt:
$1-\frac{2}{1}(\frac{1}{4})+\frac{2.4}{1.2}(\frac{1}{ 4})^2-\frac{2.4.6}{1.2.3}(\frac{1}{4})^3-...$
General term: $\frac{(1)(2)(3)...(r)}{r!}(-\frac{1}{2})^r$
Which is the general term of the series $(1+x)^n$ where $n=1$ and $x=\frac{1}{2}$
So the sum $=(1+(-\frac{1}{2}))^1=\frac{1}{2}$
Answer is: $\frac{2}{3}$

Thanks!

**TheEmptySet** · July 29th 2010, 06:59 PM

Consider the series

$\displaystyle f(x)=2\sum_{n=0}^{\infty}{x}^{2n-2}=\frac{2}{x^2(1-x^2)}$

Now integrate to get

$\displaystyle g(x)=\int f(x)dx=2\sum_{n=0}^{\infty}\frac{1}{2n-1}{x}^{2n-1}=\int\frac{2}{x^2(1-x^2)}dx$

Note that this is your series when $x=\frac{1}{3}$

**Renji Rodrigo** · July 29th 2010, 07:24 PM

Other solution ( using D for derivative and arctgh(x) for ArcTanh(x) )

we know that $D arctgh (x) =\frac{1}{1-x^{2}}$ and $arctgh (0)=0$ so

$\int^{x}_{0}\frac{1}{1-y^{2}}dy=arctgh(x)$

using the geometric series for $\frac{1}{1-y^{2}}$ we have
$\int^{x}_{0}\frac{1}{1-y^{2}}dy=\int^{x}_{0}\sum\limits^{\infty}_{k=0}y^{ 2k}dy= \sum\limits^{\infty}_{k=0}\int^{x}_{0}y^{2k}dy=\su m\limits^{\infty}_{k=0}\frac{x^{2k+1}}{2k+1}$

so $arctgh (x)=\sum\limits^{\infty}_{k=0}\frac{x^{2k+1}}{2k+1 }.$
taking x=1/3

$2arctgh (\frac{1}{3})=2\sum\limits^{\infty}_{k=0}\frac{ (1/3)^{2k+1}}{2k+1}.$

**arze** · July 29th 2010, 07:38 PM

So the sum would be $\int\frac{2}{x^2(1-x^2)}dx$ when $x=\frac{1}{3}$ ?
$\int^{\frac{1}{3}}_0 \frac{2}{x^2(1-x^2)}dx=[-\frac{2}{x}-\ln(1-x)+\ln(x+1)]^{\frac{1}{3}}_0=-5.31$ to three s.f.
answer is $\ln 2$

**Renji Rodrigo** · July 30th 2010, 08:58 AM

Another try

$\sum\limits^{\infty}_{k=0}\frac{1}{a^{2k}(2k+1)}=\ sum\limits^{\infty}_{k=0}\frac{1}{a^{2k}}\int^{1}_ {0}x^{2k}=\int^{1}_{0}\sum\limits^{\infty}_{k=0}\b igg(\frac{x^{2}}{a^{2}}\bigg)^{k}dx= \int^{1}_{0}\frac{a^{2}}{a^{2}-x^{2}}dx$

by the parcial fraction decomposition
$\frac{a^{2}}{a^{2}-x^{2}}=\frac{a}{2}\bigg(\frac{1}{a+x}-\frac{1}{x-a} \bigg)$

we get
$2\sum\limits^{\infty}_{k=0}\frac{1}{a^{2k+1}(2k+1) }=\ln (\frac{a+1}{a-1})$
taking a=3

$2\sum\limits^{\infty}_{k=0}\frac{1}{3^{2k+1}(2k+1) }=\ln (\frac{4}{2})= \ln (2)$

**Opalg** · July 30th 2010, 12:45 PM

Originally Posted by arze

$1-\frac{2}{4}+\frac{2.4}{4.8}-\frac{2.4.6}{4.8.12}-...$
My attempt:
$1-\frac{2}{1}(\frac{1}{4})+\frac{2.4}{1.2}(\frac{1}{ 4})^2-\frac{2.4.6}{1.2.3}(\frac{1}{4})^3-...$
General term: $\frac{(1)(2)(3)...(r)}{r!}(-\frac{1}{2})^r$ (Correct.)
Which is the general term of the series $(1+x)^n$ where $n=1$ (Should be n = –1.) and $x=\frac{1}{2}$
So the sum $=(1+(-\frac{1}{2}))^1=\frac{1}{2}$ (Make that $(1+\frac{1}{2})^{-1}=\frac23$ .)
Answer is: $\frac{2}{3}$

..

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