Question

If $T_n=\frac{(2n+1)\left(2n^2+2n-1\right)}{(n-1{)}^2{n}^2(n+1{)}^2(n+2{)}^2},n\in N,n\ge 2$, then $\underset{n\to \infty }{\mathrm{Lim}}displaystyle{\sum }_{r=2}^n{T}_r$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{9}$

Answer 1

Answer: (D)

$T_r=\frac{(2r+1)\left(2r^2+2r-1\right)}{(r-1{)}^2{r}^2(r+1{)}^2(r+2{)}^2}=\frac{(2r+1)\left(2r^2+2r-1\right)}{{\left(r^2-1\right)}^2{\left(r^2+2r\right)}^2}$
$=\frac{(2r+1)\left(2r^2+2r-1\right)}{{\left(r^2-1\right)}^2{\left((r+1{)}^2-1\right)}^2}=\frac{1}{{\left(r^2-1\right)}^2}-\frac{1}{{\left((r+1{)}^2-1\right)}^2}$
Now, $\sum _{r=2}^n{T}_r=\left(\frac{1}{{\left(2^2-1\right)}^2}-\frac{1}{{\left(3^2-1\right)}^2}\right)+\left(\frac{1}{{\left(3^2-1\right)}^2}-\frac{1}{{\left(4^2-1\right)}^2}\right)+\dots \dots +\left(\frac{1}{{\left(n^2-1\right)}^2}-\frac{1}{(n+1{)}^2-1}\right)$ $\therefore \underset{n\to \infty }{\mathrm{Lim}}\sum _{r=2}^n{T}_r=\frac{1}{9}$