1
$\begingroup$

How to evaluate $$\frac13+\frac13(\frac13)^3+\frac15(\frac13)^5+...$$?

I faced this particular sum in the website www.toppr.com .And it is given under the heading "Problems on Approximation"...but I cant figure out a method.How to approach?Please help!

$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

\begin{align} \sum_{n=1}^{\infty}\frac{1}{2n-1}\Big(\frac13\Big)^{2n-1}&=\sum_{n=1}^{\infty}\frac{1}{n}\Big(\frac13\Big)^{n}-\sum_{n=1}^{\infty}\frac{1}{2n}\Big(\frac13\Big)^{2n}\\ &=-\log(1-\frac13)+\frac12\log(1-\frac19)\\ &=\frac12 \log2 \end{align} where we use $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}x^{n}=-\log(1-x)$.

$\endgroup$
1
  • $\begingroup$ Alternatively, $\sum\limits_{n=0}^{\infty} \frac{1}{2n+1} x^{2n+1}=\text{arctanh }x$. $\endgroup$
    – krvolok
    Commented Aug 18, 2015 at 10:22
2
$\begingroup$

The given series is $x+\frac {x^3} 3 + \frac {x^5} 5 + \dots \big|_{x = \frac 1 3} \\ = \frac 1 2 \log \frac {1+x} {1-x} \big|_{x = \frac 1 3} \\ = \frac 1 2 \log 2 .$

$\endgroup$

You must log in to answer this question.