Question

Let $t_r=\frac{r}{1+r^2+r^4}$ then, ${lim}_{n\to \infty }{\sum }_{r=1}^n{t}_r$ equals

• A. $\frac{1}{4}$
• B. 1
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{2}$

$t_r=\frac{r}{1+r^2+r^4}$

$=\frac{r}{r^4+2r^2+1-r^2}$

$=\frac{r}{(r^2+1{)}^2-r^2}=\frac{r}{(r^2+1-r)(r^2+1+r)}$

$=\frac{1}{2}\frac{2r}{(r^2+1-r)(r^2+r+1)}$

$=\frac{1}{2}\frac{(r^2+r+1)-(r^2-r+1)}{(r^2-r+1)(r^2+r+1)}$

$=\frac{1}{2}[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}]$

$=\frac{1}{2}[\frac{1}{r^2-r+1}-\frac{1}{(r+1{)}^2-(r+1)+1}]$

$\sum _{r=1}^n{t}_r=\frac{1}{2}[\frac{1}{1-1+1}-\frac{1}{(n+1{)}^2-(n+1)+1}]$

$=\frac{1}{2}[1-\frac{1}{n^2+n+1}]=\frac{1}{2}\left[\frac{n^2+n+1-1}{n^2+n+1}\right]$

$\sum t_n=\frac{1}{2}\left[\frac{1+\frac{1}{n}}{1+\frac{1}{n}+\frac{1}{n^2}}\right]$

$\underset{n\to \infty }{lt}\sum t_r=\frac{1}{2}\left[\frac{1+0}{1+0+0}\right]$

$=\frac{1}{2}$.