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$\lim\limits_{n\to\infty}\large\frac{1^p+2^p+3^p+.....+n^p}{n^{p+1}}$ is

$\begin{array}{1 1}(a)\;\large\frac{1}{p+1}&(b)\;\large\frac{1}{1-p}\\(c)\;\large\frac{1}{p}-\frac{1}{p-1}&(d)\;\large\frac{1}{p+2}\end{array}$

1 Answer

We have $\lim\limits_{n\to\infty}\large\frac{1^p+2^p+3^p+.....+n^p}{n^{p+1}}$
$\Rightarrow \lim\limits_{n\to \infty}\sum\limits_{r=1}^n\large\frac{r^p}{n^p.n}$
$\Rightarrow \int_0^1x^pdx$
$\Rightarrow \bigg[\large\frac{x^{p+1}}{p+1}\bigg]_0^1$
$\Rightarrow \large\frac{1}{p+1}$
Hence (a) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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