lim-n-r-1-r-n-2-r-

Question Number 174630 by infinityaction last updated on 06/Aug/22

\lim_{n \to \infty} \underset{r = 1}{\sum^{\infty}} \frac{r}{n^{2} + r}

Answered by mnjuly1970 last updated on 06/Aug/22

{l i m}_{n \to \infty} {\frac{1}{1 + n^{2}} + \frac{2}{2 + n^{2}} + \frac{3}{3 + n^{2}} + \dots \frac{n}{n + n^{2}} = a_{n}}

⩽ \frac{1}{1 + n^{2}} + \frac{2}{1 + n^{2}} + \dots + \frac{n}{1 + n^{2}}

= \frac{n (n + 1)}{2 (1 + n^{2})} (1)

a_{n} ⩾ \frac{1}{n + n^{2}} + \frac{2}{n + n^{2}} + \dots + \frac{n}{n + n^{2}}

= \frac{n (n + 1)}{2 n (1 + n)}

\frac{1}{2} ⩽ a_{n} ⩽ \frac{n (n + 1)}{2 (1 + n^{2})}

{l i m}_{n \to} (a_{n}) = \frac{1}{2}

Commented by infinityaction last updated on 06/Aug/22

t h a n k s s i r

Commented by mnjuly1970 last updated on 06/Aug/22

y o u a r e w e l c o m e s i r \dots