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  4. If Tn=((2 n+1)(2 n2+2 n-1)/(n-1)2 n2(n+1)2(n+2)2), n ∈ N, n ≥ 2, then undersetn arrow ∞ textLim displaystyle∑r=2n Tr is equal to

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Q. If $T_n=\frac{(2 n+1)\left(2 n^2+2 n-1\right)}{(n-1)^2 n^2(n+1)^2(n+2)^2}, n \in N, n \geq 2$, then $\underset{n \rightarrow \infty}{\text{Lim}} displaystyle\sum_{r=2}^n T_r$ is equal to

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Solution:

$T_r=\frac{(2 r+1)\left(2 r^2+2 r-1\right)}{(r-1)^2 r^2(r+1)^2(r+2)^2}=\frac{(2 r+1)\left(2 r^2+2 r-1\right)}{\left(r^2-1\right)^2\left(r^2+2 r\right)^2}$
$=\frac{(2 r+1)\left(2 r^2+2 r-1\right)}{\left(r^2-1\right)^2\left((r+1)^2-1\right)^2}=\frac{1}{\left(r^2-1\right)^2}-\frac{1}{\left((r+1)^2-1\right)^2}$
Now, $\displaystyle\sum_{ r =2}^{ n } T _{ r }=\left(\frac{1}{\left(2^2-1\right)^2}-\frac{1}{\left(3^2-1\right)^2}\right)+\left(\frac{1}{\left(3^2-1\right)^2}-\frac{1}{\left(4^2-1\right)^2}\right)+\ldots \ldots+\left(\frac{1}{\left( n ^2-1\right)^2}-\frac{1}{( n +1)^2-1}\right)$ $\therefore \underset{n \rightarrow \infty}{\text{Lim}} \displaystyle\sum_{ r =2}^{ n } T _{ r }=\frac{1}{9}$