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find sum 1/6 + 1/24 + 1/60.....1/r^3 – r. How do we answer this question. Please answer ASAP

Kanan , 8 Years ago
Grade 11
anser 1 Answers
jagdish singh singh

Last Activity: 8 Years ago

\hspace{-0.9 cm.}$ Let $t_{r} = \frac{1}{r(r^2-1)} = \frac{1}{(r-1)r(r+1)} = \frac{1}{2}\left[\frac{1}{(r-1)r}-\frac{1}{r(r+1)}\right]$\\\\So $\sum^{n}_{r=2}t_{r} =\frac{1}{2}\sum^{n}_{r=2}\left[\frac{1}{(r-1)r}-\frac{1}{r(r+1)}\right] $\\\\Now Using Telescoping Sum, We get $\sum^{n}_{r=2}t_{r}=\frac{1}{2}\left[1-\frac{1}{n(n+1)}\right]$\\\\So we get $\sum^{n}_{r=1}t_{r} = \frac{n^2+n+1}{2n(n+1)}$ and $\lim_{n\rightarrow 0}\sum^{n}_{r=1}t_{r} = \frac{1}{2}.$

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