[gon]

น่าจะประมาณนี้ครับ

solution

$\sum_{n=1}^\infty \frac{1}{n(4n-3)} $

$= \sum_{n=1}^\infty \frac{4}{(4n-3)(4n)}$

$= \frac{4}{3} \sum_{n=1}^\infty [\frac{1}{4n-3}-\frac{1}{4n}] $

$= \frac{4}{3}\sum_{n=1}^\infty[\int_0^1 x^{4n-4} dx - \int_0^1 x^{4n-1} dx]$

$=\frac{4}{3} \int_0^1 \sum_{n=1}^\infty[x^{4n-4} - x^{4n-1}] dx$

$= \frac{4}{3}\int_0^1 [\frac{1}{1-x^4} - \frac{x^3}{1-x^4}] dx$

$=\frac{4}{3} \int_0^1 \frac{1+x+x^2}{(1+x)(1+x^2)} dx$

$=\frac{2}{3} \int_0^1 \frac{(1+x)^2 + (1+x^2) }{(1+x)(1+x^2)} dx$

$=\frac{2}{3} \int_0^1 [\frac{1+x}{1+x^2} + \frac{1}{1+x}] dx$

$= \frac{2}{3} \int_0^1 [\frac{1}{1+x^2} + \frac{x}{1+x^2} + \frac{1}{1+x}] dx$

$=\frac{2}{3}[\arctan x + \frac{1}{2}\ln(1+x^2) + \ln|1+x|] \left|\,\right._0^1 $

$=\frac{2}{3}[\frac{\pi}{4} + \frac{1}{2}\ln 2 + \ln 2 - 0 - 0 - 0]$

$=\frac{2}{3}[\frac{3}{2}\ln 2 + \frac{\pi}{4}]$

$=\ln 2 + \frac{\pi}{6}$