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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Evaluate $ \lim_{n \rightarrow \infty} \bigg[ \large\frac{1^p + 2^p + 3^p....+ n^p}{n^{p+1}} \bigg ] $, where $p \gt -1$

(A) $p+1 \quad$ (B)$ \large\frac{1}{p-1} \quad$ (C) $\large\frac{1}{p+1} \quad$ (D) $p-1$

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$\lim_{n \rightarrow \infty} \bigg[ \large\frac{1^p + 2^p + 3^p....+ n^p}{n^{p+1}} \bigg ] = \large\frac{1}{n}$$ \lim_{n \rightarrow \infty} \bigg[ \large\frac{1^p + 2^p + 3^p....+ n^p}{n^p} \bigg ] $
$\quad = \large\frac{1}{n}$$ \lim_{n \rightarrow \infty} \bigg[ \bigg(\frac{1}{n}\bigg)^p + \bigg(\frac{2}{n}\bigg)^p + \bigg(\frac{3}{n}\bigg)^p + ... + \bigg(\frac{n}{n}\bigg)^p \bigg]$
Let $\large\frac{1}{n} $$ = h \rightarrow nh = 1 $ and as $n \rightarrow \infty$, $h \rightarrow 0$
$ \large\frac{1}{n}$$ \lim_{n \rightarrow \infty} \bigg[ \bigg(\frac{1}{n}\bigg)^p + \bigg(\frac{2}{n}\bigg)^p + \bigg(\frac{3}{n}\bigg)^p + ... + \bigg(\frac{n}{n}\bigg)^p \bigg]$ $ = \lim_{h \rightarrow 0} \bigg[ (1h)^p + (2h)^p + (3h)^p...+ (nh)^p \bigg]$
$ \quad = \lim_{h \rightarrow 0} h \large \sum_{r=1}^{n}$$ (rh)^p$$ = \Large \int_{\normalsize 0}^{\normalsize 1}$$ x^p dx$
$\quad = \bigg[ \large\frac{x^p + 1}{p+1} \bigg]_{0}^{1}$
$\quad = \large\frac{1}{p+1}$ (when $p \gt -1$, this evaluates to $ \large\frac{1}{p+1} $$- 0$)
answered Mar 20, 2014 by balaji.thirumalai
 

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