# Prove that $\frac{1}{1}-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\ldots= \frac{1}{3} \left( {\frac{\pi}{\sqrt{3}}+ \log 2} \right)$

How do I show that

$$\frac{1}{1}-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\ldots= \frac{1}{3} \left( {\frac{\pi}{\sqrt{3}}+ \log 2} \right)?$$

This problem belongs to Riemann Theory of Definite Integral, and not to any series summation. I recommend an answer which is to the topic i.e., Riemann Theory of D.I..

Thanks!

• Please include the question in the body and not just in the title. Sep 7, 2011 at 9:43

HINT: First argue out the convergence by alternating series test. Then consider $f(t) = 1 - t^3 + t^6 - \cdots$ where $0 \leq t \lt 1$. Integrate $f(t)$ in two different ways to get to the answer.
• Yes! I got it. I integrated $\dfrac{1}{1+t^3}$ (result: $\dfrac{1}{3} \left({\frac{\pi}{\sqrt {3}}+\log 2} \right)$ ) which also equals to {your} $f(t)= 1-t^3+t^6-\ldots$ (result: L.H.S.) . [Sorry for late response. I was busy in solving other problems. Many thanks to you.] Sep 7, 2011 at 11:43