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This is the question $$S=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...\infty $$ I found the $T_n=\frac{3}{n\left(n+1\right)}$. Could anyone please help me solve rest of the question? I was hoping to solve this expression $S_n=\sum _{n=1}^{\infty }\:\frac{3}{n\left(n+1\right)}$ I entered this expression onto symbolab, it gave an explanation using 'telescoping series test' which I am not familiar with.

Please be free to use other methods if they are shorter to solve this question. I just need an explanation.

Anonymousstriker38596
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  • Just a notational point: you shouldn't include $\infty$ at the end of your infinite summations. I get that you're trying to indicate that there are infinitely many terms of the sum, but leaving the $\ldots$ by itself does that already. Besides, putting the number of terms at the end makes as much sense as denoting $\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^n}$ by $\frac{1}{2}+\frac{1}{2^2}+\ldots+n$. – Theo Bendit Mar 17 '22 at 09:02
  • Ohk thanks, looks like the booklet made an error. Is it possible to solve this question without using telescoping series? I am not familiar with that method. – Anonymousstriker38596 Mar 17 '22 at 09:04
  • I'm not exactly sure what the question is (is it about convergence, or computing the exact value?). If you look at the top answer in the dupe target, you'll find an answer that doesn't rely on any series tests. The nice thing about telescoping series is that computing the partial sums are easy, and you don't really need anything more than the definition of series convergence to tell that they converge. – Theo Bendit Mar 17 '22 at 09:12
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    I just googled telescoping series and found a video about it, 2 mins into it I realised that we had been taught this in class but a name for the method wasn't mentioned. Thanks Theo, I got it. It was not that hard. I just had to rewrite the expression as $S_n=\sum _{n=1}^{\infty }:\frac{3}{n\left(n+1\right)}=:3\sum _{n=1}^{\infty }:\frac{\left(n+1\right)-n}{n\left(n+1\right):}$ then put the values 1, 2, 3 find a pattern as to what cancels and get the final ans. – Anonymousstriker38596 Mar 17 '22 at 09:20
  • Ah, mystery solved! – Theo Bendit Mar 17 '22 at 09:27

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