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

Question

If $\sum_\limits{r=1}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}$, then $\lim _\limits{n \rightarrow \infty} \sum_\limits{r=1}^n\left(\frac{1}{T_r}\right)$ is equal to :

Accepted Answer

$\begin{aligned} & T_n=S_n-S_{n-1} \\ & \Rightarrow T_n=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3) \\ & \Rightarrow \frac{1}{T_{\mathrm{n}}}=\frac{8}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\ & \lim _{\mathrm{n} ightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{T_{\mathrm{r}}}=\lim _{\mathrm{n} ightarrow \infty} 8 \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\ & =\lim _{\mathrm{n} ightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)}-\frac{1}{(2 \mathrm{n}+1)(2 \mathrm{n}+3)}ight) \\ & =\lim _{\mathrm{n} ightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}ight)+\left(\frac{1}{3.5}-\frac{1}{5.7}ight)+\ldotsight] \\ & =\frac{2}{3}\end{aligned}$

Solutions · 1