Question

Find the sum of infinity terms of the series 1-1/2+1/4-1/8+1/10-1/12 and so on up infinity​

Answer 1

$a=1,r=-\frac{1}{2}$

.

$\tt\red{\therefore S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{1 \times\left[1-\left(-\frac{1}{2}\right)^{n}\right]}{1-\left(-\frac{1}{2}\right)}=\frac{\left[1-\left(-\frac{1}{2}\right)^{n}\right]}{\frac{3}{2}}=\frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^{n}\right]}$

.

$\tt\pink{\Rightarrow S_{\infty}=\underset{n \rightarrow \infty}{L t} \frac{2}{3}\left[1-\left(\frac{-1}{2}\right)^{n}\right]=\frac{2}{3}(1-0)=\frac{2}{3}}$

Note. We could straightway use the result $\tt\left[S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\left(-\frac{1}{2}\right)}=\frac{1}{\frac{3}{2}}=\frac{2}{3}\right]$