Find-lim-n-n-1-2n-1-2-1-2n-3-2-1-4n-1-2-

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}} + \dots + \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}} + \dots . + \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}} + \dots + \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

t h a n k y o u v e r y m u c h

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