nice-calculus-calculate-lim-n-k-1-n-k-2n-2-k-

Question Number 130845 by mnjuly1970 last updated on 29/Jan/21

\dots n i c e c a l c u l u s \dots

c a l c u l a t e ::

{l i m}_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{k}{2 n^{2} + k} = ?

Answered by Ar Brandon last updated on 29/Jan/21

1 ⩽ k ⩽ n \Rightarrow {2 n}^{2} + 1 ⩽ {2 n}^{2} + k ⩽ {2 n}^{2} + n

\Rightarrow \frac{k}{{2 n}^{2} + n} ⩽ \frac{k}{{2 n}^{2} + k} ⩽ \frac{k}{{2 n}^{2} + 1}

\Rightarrow \frac{n (n + 1)}{2 ({2 n}^{2} + n)} ⩽ \underset{k = 1}{\sum^{n}} \frac{k}{{2 n}^{2} + k} ⩽ \frac{n (n + 1)}{2 ({2 n}^{2} + 1)}

\Rightarrow \frac{1}{4} ⩽ \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{k}{{2 n}^{2} + k} ⩽ \frac{1}{4}

Commented by mnjuly1970 last updated on 29/Jan/21

h i m a s t e r b r a n d o n a n s w e r i s \frac{1}{4}

Commented by mnjuly1970 last updated on 29/Jan/21

i n t h i r d l i n e \frac{n (n + 1)}{2 (2 n^{2} + n)} ⩽ A_{n} ⩽ \frac{n (n + 1)}{2 (2 n^{2} + 1)}

y o u r a n s w e r i s c o r r e c t

{l i m}_{n \to \infty} A_{n} = \frac{1}{4}

Commented by Ar Brandon last updated on 29/Jan/21

Oh! thanks for remark, Sir