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Q.
Find the sum of the infinite series $\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\frac{1}{63}+\ldots \ldots$
Sequences and Series
Solution:
$T _{ n }=\frac{1}{3}\left[\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\ldots \ldots ..\right]=\frac{1}{3}\left[\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\ldots \ldots . ..\right]$
$\therefore T _{ n }=\frac{1}{3}\left[\frac{1}{1+2+3+\ldots \ldots 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: $T _{ n }=\frac{1}{3}\left[\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\ldots \ldots \ldots\right]$
$=\frac{2}{3}\left[\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\ldots \ldots . . .\right]$
hence $T_n$ using method of diff; $T_n=\frac{2}{3} \frac{1}{(n+1)(n+2)}=\frac{2}{3}\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$
$\therefore S _{\infty}=\frac{2}{3} \cdot \frac{1}{2}=\frac{1}{3}$