How to sum this series :
$\frac{1}{9} + \frac{1}{18}+\frac{1}{30}+\frac{1}{45} + \frac{1}{65}......$
I am not getting any clue only a hint will be suffice please help. thanks..
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1$\begingroup$ Let $I=\frac{1}{9} + \frac{1}{18}+\frac{1}{30}+\frac{1}{45} + ......$ then $3I=\frac{1}{3} + \frac{1}{6}+\frac{1}{10}+\frac{1}{15} + ......$ $\endgroup$– mikeJun 20, 2014 at 3:42
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4$\begingroup$ Is the last term really $\frac{1}{65}$ instead of $\frac{1}{63}$? $\endgroup$– achille huiJun 20, 2014 at 8:26
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$\begingroup$ yes it is $\frac{1}{65}$ and not $\frac{1}{63}$ $\endgroup$– user108258Jun 20, 2014 at 13:25
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$\begingroup$ Are there any more terms to the sequence? $\endgroup$– Mathmo123Jun 20, 2014 at 13:29
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$\begingroup$ no there is no more term in the sequence $\endgroup$– user108258Jun 20, 2014 at 16:51
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2 Answers
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Observe \begin{align*} \frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\ldots&=\frac{1}{3+3\cdot2}+\frac{1}{3+3\cdot2+3\cdot3}+\frac{1}{3+3\cdot2+3\cdot3+3\cdot4}+\ldots\\ &=\sum_{k=2}^{\infty}{\frac{1}{3\sum_{j=1}^{k}{j}}}\\ &=\sum_{k=2}^{\infty}{\frac{1}{3\left[\frac{k(k+1)}{2}\right]}}\\ &=\sum_{k=2}^{\infty}{\frac{2}{3\left[k(k+1)\right]}}\\ &=\frac{2}{3}\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\ \end{align*} Last one is a telescopic series, note we have $$\frac{2}{3}\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{2}{3}\left(\frac{1}{2}\right)=\frac{1}{3}$$
answered Jun 20, 2014 at 4:04
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$\begingroup$ The answer should be ⅓ I think. I think you shouldn't have the $- 1$ on the third line. Also - the /2 is wrong in that line. $\endgroup$ Jun 20, 2014 at 4:06
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$\begingroup$ $\sum_{k=2}^{\infty}{\frac{1}{3\sum_{j=2}^{k}{j}}}$ should be $\sum_{k=2}^{\infty}{\frac{1}{3\sum_{j=1}^{k}{j}}}$ in the second line. $\endgroup$ Jun 20, 2014 at 4:06
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1$\begingroup$ Thanks @triple_sec, you are right, I was wrong. Already I have corrected it. $\endgroup$ Jun 20, 2014 at 4:10
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This series can be computed as $$\frac{1}{3}\sum_{k=0}^{\infty}\frac{1}{(\sum_{\ell=0}^k\ell)+2k+3}.$$ The partial sums are $$\frac{1}{3}\sum_{k=0}^{K}\frac{1}{(\sum_{\ell=0}^k\ell)+2k+3}=\frac{K+1}{3(K+3)},$$ and the limit is $1/3$. (Check for the first few values of $k$ and $K$ to convince yourself that the formulae are indeed correct and proceed by induction. Start with a value of $k=0$, not $1$. Hint: $\sum_{\ell=0}^k\ell=k(k+1)/2$.)
