lim-x-0-r-1-2n-1-r-n-please-help-

Question Number 50293 by pooja24 last updated on 15/Dec/18

l i \underset{x \to 0}{m} \underset{r = 1}{\sum^{2 n}} (\frac{1}{r + n}) = ?

p l e a s e h e l p

Commented by maxmathsup by imad last updated on 15/Dec/18

l e t S_{n} = \sum_{k = 1}^{2 n} \frac{1}{k + n} \Rightarrow S_{n} = \sum_{k = 1}^{n} \frac{1}{k + n} + \sum_{k = n + 1}^{2 n} \frac{1}{k + n} = A_{n} + B_{n}

{w e h a v e A_{n} = \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k}{n}} \to_{n \to + \infty} \int_{0}^{1} \frac{d x}{1 + x} = l n ∣ 1 + x ∣]}_{0}^{1} = l n (2)

c h a n g e m e n t k - n = p g i v e B_{n} = \sum_{p = 1}^{n} \frac{1}{n + p + n} = \sum_{p = 1}^{n} \frac{1}{p + 2 n}

= \frac{1}{n} \sum_{p = 1}^{n} \frac{1}{2 + \frac{p}{n}} \to_{n \to + \infty} \int_{0}^{1} \frac{d x}{2 + x} = {[l n ∣ 2 + x ∣]}_{0}^{1} = l n (x) - l n (2) \Rightarrow

{l i m}_{n \to + \infty} S_{n} = {l i m}_{n \to + \infty} (A_{n} + B_{n}) = l n (2) + l n (3) - l n (2) = l n (3)

Commented by maxmathsup by imad last updated on 15/Dec/18

\int_{0}^{1} \frac{d x}{2 + x} = l n (3) - l n (2) .

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

t h e r e i s n o \infty i n q u e s t i o n \dots

Commented by maxmathsup by imad last updated on 15/Dec/18

i t h i n k t h e Q u e s t o n i s f i n d {l i m}_{n \to + \infty} \sum_{r = 1}^{2 n} \frac{1}{r + n}

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

T_{r} = \frac{1}{r + n}

T_{1} = \frac{1}{1 + n}

T_{2} = \frac{1}{2 + n}

. .

\dots

T_{2 n} = \frac{1}{2 n + n}

S_{2 n} = \frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n + n}

n o w

\frac{1}{1 + n} < \frac{1}{n}

\frac{1}{2 + n} < \frac{1}{n}

\dots

\dots

\frac{1}{2 n + n} < \frac{1}{n}

s o S_{2 n} < 2 n \times \frac{1}{n}

S_{2 n} < 2

i n t h e p r o b l e m n o m e n t i o n o f x, s o p l s c h e c k

t h e q u e s t i o n \dots

S_{2 n} = \underset{r = 1}{\sum^{2 n}} \frac{1}{r + n} < 2

\lim_{x \to 0}

Answered by ajfour last updated on 15/Dec/18

\lim_{n \to \infty} \underset{r = 1}{\sum^{2 n}} \frac{1}{r + n} = \int_{0}^{2} \frac{d x}{x + 1} = \ln 3 .

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

t h e r e i s n o \infty i n q u e s t i o n \dots