Solutions - 0240-08

(a)

Recall the series for ${\tan }^{-1}$:

$${\tan }^{-1}(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n+1}}{2n+1}$$

This matches our series if we set $x={\displaystyle \frac{\sqrt{3}}{3}}$. So the total sum is:

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^n\frac{{\sqrt{3}}^{2n+1}}{3^{2n+1}(2n+1)}\gg \gg {\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\\ \\ \gg \gg {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\gg \gg \frac{\pi }{6}\end{array}$$

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(b)

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^{n+1}\frac{1}{5^{n+1}(n+1)}\gg \gg \sum _{n=0}^{\infty }\frac{{(-1/5)}^{n+1}}{n+1}\\ \\ \gg \gg -\sum _{n=0}^{\infty }-\frac{{(-1/5)}^{n+1}}{n+1}\gg \gg {-\sum _{n=0}^{\infty }-\frac{x^{n+1}}{n+1}|}_{x=-1/5}\\ \\ \gg \gg -\ln (1-x){|}_{x=-1/5}\gg \gg -\ln (1+1/5)\\ \\ \gg \gg -\ln (6/5)\gg \gg \ln (5/6)\end{array}$$