The sum of the infinite series frac{1}{9} + f | vedclass

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Question

(A) The given series is $S = \frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + ..........\infty$. We can rewrite the terms as

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$\frac{1}{3}$

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(A) The given series is $S = \frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + ..........\infty$. We can rewrite the terms as $S = \frac{1}{3} (\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..........\infty)$. Multiplying by $2$,we get $2S = \frac{2}{3} (\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..........\infty)$. Note that the denominators are triangular numbers $T_n = \frac{n(n+1)}{2}$. Thus,the series inside is $\sum_{n=2}^{\infty} \frac{2}{n(n+1)} = 2 \sum_{n=2}^{\infty} (\frac{1}{n} - \frac{1}{n+1})$. This is a telescoping series: $2 [(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + ..........] = 2 (\frac{1}{2}) = 1$. Therefore,$2S = \frac{2}{3} (1)$,which gives $S = \frac{1}{3}$.

The sum of the infinite series $\frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + ..........\infty$ is equal to :-

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