Find lim_(n→∞) (n((1/((2n+1)^2 )) + (1/((2n+3)^2 )) + ... + (1/((4n−1)^2 ))))


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Question Number 62571 by Joel122 last updated on 23/Jun/19

$Find$ $\lim_{n \to \infty} (n (\frac{1}{{(2 n + 1)}^{2}} + \frac{1}{{(2 n + 3)}^{2}} + . . . + \frac{1}{{(4 n - 1)}^{2}}))$
Commented by Prithwish sen last updated on 23/Jun/19

$li \underset{n \to \infty}{m} \frac{1}{n} [\frac{l}{{(2 + \frac{1}{n})}^{2}} + \frac{1}{{(2 + \frac{3}{n})}^{2}} + . . . . + \frac{1}{{(2 + \frac{2 n - 1}{n})}^{2}}]$ $= \lim_{n \to \infty} \frac{1}{n} \underset{r = 1}{\sum^{n}} \frac{1}{{(2 + \frac{2 r - 1}{n})}^{2}}$ $Putn = \frac{1}{h} n \to \infty \Rightarrow h \to 0$ $= \lim_{h \to 0} h \underset{r = 1}{\sum^{n}} \frac{1}{{(2 + 2 rh)}^{2}}$ $= \underset{0}{\int^{1}} \frac{dx}{{(2 + 2 x)}^{2}} ∵ nh = 1 = 1 - 0$ $= \frac{1}{4} {[- \frac{1}{(1 + x)}]}_{0}^{1}$ $= \frac{1}{4} [- \frac{1}{2} + 1]$ $= \frac{1}{8}$
Commented by Joel122 last updated on 23/Jun/19

$The answer is \frac{1}{8}$
Commented by Tony Lin last updated on 23/Jun/19

$= \lim_{n \to \infty} (\frac{1}{2} \times \frac{2}{n} (\frac{1}{{(2 + \frac{1}{n})}^{2}} + \frac{1}{{(2 + \frac{3}{n})}^{2}} + . . . + \frac{1}{{(4 - \frac{1}{n})}^{2}}))$ $= \frac{1}{2} \int_{2}^{4} \frac{1}{x^{2}} d x$ $= \frac{1}{2} {[- \frac{1}{x}]}_{2}^{4}$ $= - \frac{1}{8} + \frac{1}{4}$ $= \frac{1}{8}$
Commented by JDamian last updated on 23/Jun/19

$P l e a s e, w o u l d y o u m i n d t o e x p l a i n h o w t h e s u m$ $o f a d i s c r e t e s u c e s s i o n t u r n s i n t o a n i n t e g r a l$ $i n ℜ ?$
Commented by Tony Lin last updated on 23/Jun/19

$T h e d e f i n i t e i n t e g r a l o f a c o n t i n u o u s$ $f u n c t i o n f (x) o v e r t h e i n t e r v a l [a, b]$ $, d e n o t e d b y \int_{a}^{b} f (x) d x, i s t h e l i m i t o f$ $R i e m a n n s u m a s t h e n u m b e r o f$ $s u b d i v i s i o n s a p p r o a c h e s i n f i n i t y$ $i . e \int_{a}^{b} f (x) d x = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} △ x f (x_{k})$ $w h e r e △ x = \frac{b - a}{n} a n d x_{k} = a + k △ x$ $n o w w e h a v e △ x = \frac{2}{n}, m e a n s t h e$ $c o n s t a n t w i d t h o f t h e r e c t a n g l e,$ $a n d x_{k} i s t h e x v a l u e o f t h e r i g h t$ $e d g e o f t h e k^{t h} r e c t a n g l e, f r o m$ $x_{1} = 2 + \frac{1}{n} t o x_{n} = 4 - \frac{1}{n}, t h e n f (x_{k})$ $w i l l g i v e u s t h e h e i g h t o f e a c h$ $r e c t a n g l e, a n d h e r e f (x) = \frac{1}{x^{2}}, s o t h e$ $a r e a o f t h e k^{t h} r e c t a n g l e i s$ $\frac{2}{n} \times \frac{1}{{(2 + \frac{2}{n} k)}^{2}}, a n d w e s u m t h a t f o r$ $v a l u e s o f k f r o m 1 t o n$ $R (n) = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{{(2 + \frac{2}{n} k)}^{2}} \times \frac{2}{n}$ $n o w w e c a n r e p r e s e n t t h e a c t u a l$ $a r e a a s a l i m i t :$ $\int_{2}^{4} \frac{1}{x^{2}} d x = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{{(2 + \frac{2}{n} k)}^{2}} \times \frac{2}{n} = \frac{1}{4}$ $b u t n o w i t g i v e s \frac{1}{n} n o t \frac{2}{n},$ $s o i t n e e d s t o b e m u l t i p l i e d b y \frac{1}{2}$
Commented by Joel122 last updated on 24/Jun/19

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