lim_(x→0) Σ_(r=1) ^(2n) ((1/(r+n)))=? please help


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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

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

	$T_{r} = \frac{1}{r + n}$ $T_{1} = \frac{1}{1 + n}$ $T_{2} = \frac{1}{2 + n}$ $. .$ $. . .$ $T_{2 n} = \frac{1}{2 n + n}$ $S_{2 n} = \frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + . . . + \frac{1}{2 n + n}$ $n o w$ $\frac{1}{1 + n} < \frac{1}{n}$ $\frac{1}{2 + n} < \frac{1}{n}$ $. . .$ $. . .$ $\frac{1}{2 n + n} < \frac{1}{n}$ $s o S_{2 n} < 2 n \times \frac{1}{n}$ $S_{2 n} < 2$ $i n t h e p r o b l e m n o m e n t i o n o f x, s o p l s c h e c k$ $t h e q u e s t i o n . . .$ $S_{2 n} = \underset{r = 1}{\sum^{2 n}} \frac{1}{r + n} < 2$ $\lim_{x \to 0}$
	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

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