An infinite series

[akotronis]

Calculate the following:

\(\displaystyle1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots\).

[Demetres]

Let us write \(a_n\) for the \(n\)-th term of the series and \(s_n\) for the partial sum of the first \(n\) terms. Observe that \[ s_{6n} = \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{8n-1} \right) - \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{4n} \right) = H_{8n} - \frac{1}{2}H_{4n} - \frac{1}{2}H_{2n}\] where \(H_m\) is the partial sum of the first \(m\) terms of the Harmonic series. Since \(H_m = \log{m} + \gamma + o(1)\) we get that \[ s_{6n} \sim \log(8n) + \gamma - \frac{1}{2}\log(4n) - \frac{\gamma}{2} - \frac{1}{2}\log(2n) - \frac{\gamma}{2} + o(1) = \frac{1}{2}\log{8} + o(1).\] Since also \(a_n = o(1)\) it follows that the series converges to \(\frac{1}{2}\log{8}.\)

[akotronis]

Thank you. We can also write

\(\displaystyle s_n=\sum_{k=0}^{n}\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2}=H_{4n+3}-\frac{1}{2}\left(H_{2n+1}+H_{n+1}\right)\to\frac{3\ln2}{2}\).

Two more approaches can be seen here .