Question

Let $T_n=\sum _{r=1}^n\frac{n}{r^2-2r.n+2n^2},S_n=\sum _{r=0}^{n-1}\frac{n}{r^2-2r.n+{2n}^2}$, then

• A. $T_n>S_n\forall nϵN$
• B. $T_n>\frac{\pi }{4}$
• C. $S_n<\frac{\pi }{4}$
• D. $\underset{n-\infty }{lim}{S}_n=\frac{\pi }{4}$

Answer 1

The correct option is A $T_n>S_n\forall nϵN$

Sol $T_n=\sum _{r=1}^n\frac{n}{r^2-2r.n+2n^2},S_n=\sum _{r=0}^{n-1}\frac{n}{r^2-2r.n+{2n}^2}$

On expanding $T_n$ and $S_2$ we get

$T_n=\frac{n}{2n^2-2n+1}+\frac{n}{2n^2-4n+4}+\frac{n}{2n^2-6n+9}+...+\frac{n}{n^2}...\left(1\right)$

and

$S_n=\frac{n}{2n^2}+\frac{n}{2n^2-2n+1}+\frac{n}{2n^2-4n+4}+...+\frac{n}{2n^1-2n^2+2n+(n-1{)}^2}$

$S_n=\frac{n}{2n^2}+\frac{n}{2n^2-2n+1}+\frac{n}{2n^2-2n+4}+...+\frac{n}{n^2+1}...\left(2\right)$

from eq (1)

on comparing $T_n$ & $S_2$ we can clearly

see that

$T_n\Rightarrow S_n$

as $T_n=P+\frac{1}{n}$

and $S_n=P+\frac{1}{2n}$

as $P=$ common part in eq (1) & (2)

to $[T_n>S_n\forall nϵN]$

Hence from equation (1) & (2)

$[T_n=S_n+\frac{1}{2n}]$