If _r=1^nT_r=((2n-1)(2n+1)(2n+3)(2n+5))/(64), then n lim _r=1^n(1T_r) is equal to :

Question

If $\sum _{r=1}^{n}{T}_{r}=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\underset{n	o \infty}{lim}\sum _{r=1}^{n}(\frac{1}{{T}_{r}})$ is equal to :

Accepted Answer

$\begin{gathered} {T}_{n}={S}_{n}-{S}_{n-1} \ \Rightarrow {T}_{n}=\frac{1}{8}(2n-1)(2n+1)(2n+3) \ \Rightarrow \frac{1}{{T}_{n}}=\frac{8}{(2n-1)(2n+1)(2n+3)} \ \underset{n	o \infty}{lim}\sum _{r=1}^{n}\frac{1}{{T}_{r}}=\underset{n	o \infty}{lim}8\sum _{r=1}^{n}\frac{1}{(2n-1)(2n+1)(2n+3)} \ =\underset{n	o \infty}{lim}\frac{8}{4}\sum (\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}) \ =\underset{n	o \infty}{lim}2[(\frac{1}{1.3}-\frac{1}{3.5})+(\frac{1}{3.5}-\frac{1}{5.7})+\ldots ] \ =\frac{2}{3} \end{gathered}$

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