Question

Find the sum of the infinite series $\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\frac{1}{63}+\dots \dots$

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

$T_n=\frac{1}{3}\left[\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\dots \dots ..\right]=\frac{1}{3}\left[\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\dots \dots ...\right]$
$\therefore T_n=\frac{1}{3}\left[\frac{1}{1+2+3+\dots \dots n+(n+1)}\right]=\frac{1}{3}\left[\frac{2}{(n+1)(n+2)}\right]=\frac{2}{3}\left[\frac{1}{(n+1)}-\frac{1}{(n+2)}\right]$
Now proceed.
Aliter: $T_n=\frac{1}{3}\left[\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\dots \dots \dots \right]$
$=\frac{2}{3}\left[\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\dots \dots ...\right]$
hence $T_n$ using method of diff; $T_n=\frac{2}{3}\frac{1}{(n+1)(n+2)}=\frac{2}{3}\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$
$\therefore S_{\infty }=\frac{2}{3}\cdot \frac{1}{2}=\frac{1}{3}$