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I was asked to sum the given series:

  • $\displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n\cdot (2n-1)}=\frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty$

Here i workout the details.

\begin{align*} \sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n-1)} &= \frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty \\ &= 1 + \Bigl( \frac{1}{2} - \frac{1}{3}\Bigr) + \frac{1}{2} \Bigl(\frac{1}{3} - \frac{1}{5}\Bigr) + \frac{1}{3} \Bigl(\frac{1}{4}-\frac{1}{7}\Bigr) + \cdots \infty \\ &= 1 + \biggl[ \frac{1}{1 \cdot 2 } + \frac{1}{2 \cdot 3} + \cdots \infty\biggr] - \biggl[ \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 5} + \frac{1}{3 \cdot 7} + \cdots \infty \biggr] \\ &= 2 - \sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n+1)} \end{align*}

Now for summing that second sum note that $$-\log(1-x^{2}) =-\biggl[x^{2} + \frac{x^{4}}{2} + \frac{x^{6}}{3} + \cdots \infty\biggr]$$

Have i done this correctly or is there some error in computing the sum. Now what i need to evaluate is the integral $$\int\limits_{0}^{1} \log(1-x^{2}) \ \rm{dx}$$

Can anyone tell me how to evaluate this integral. I tried by using Integration by parts but that didn't actually work out. Moreover, i would also be interested to see if there are other ways of solving this sum.

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1 Answer 1

up vote 9 down vote accepted

We have, $$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)} = 2 \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} \right)$$

The RHS has the alternating harmonic series, and its value is $\ln 2$.

So your final answer would be $2 \ln 2$

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