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Evaluate: $\displaystyle \lim_{n \rightarrow \infty}\left(\frac {(n+1)^m+(n+2)^m+\cdots+(n+k)^m}{n^{m-1}}-kn\right)$ ,where $m$ and $k$ are fixed positive integers

MY ATTEMPT:$\displaystyle \lim_{n \rightarrow \infty}\left(\frac {(n+1)^m+(n+2)^m+\cdots+(n+k)^m}{n^{m-1}}-kn\right)$ $=\displaystyle \lim_{n \rightarrow \infty}\left(\frac {(1+\frac 1 n)^m+(1+\frac 2 n)^m+\cdots+(1+\frac k n)^m}{n^{-1}}-kn\right)$

$=\displaystyle \lim_{n \rightarrow \infty}\sum_{i=1}^k\left(\left(1+\frac i n\right)^m-k\right)\frac 1 n$.

Here i don't know how to apply reimann sum and what will be the limit of integration.

Note:Answer of the problem is $\frac{k(k+1)}{2}m$

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2 Answers 2

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HINT:

$$\sum_{r=1}^k(n+r)^m=kn^m+\binom m1n^{m-1}\sum_{r=1}^kr+O(n^{m-2})$$

$$\frac {(n+1)^m+(n+2)^m+\cdots+(n+k)^m}{n^{m-1}}-kn$$ $$=\dfrac{\binom m1n^{m-1}\sum_{r=1}^kr+O(n^{m-2})}{n^{m-1}}$$ $$=\binom m1\sum_{r=1}^kr+O\left(\dfrac1n\right)$$

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$=\displaystyle \lim_{n \rightarrow \infty} n\left({(1+\frac 1 n)^m+(1+\frac 2 n)^m+\cdots+(1+\frac k n)^m}-k\right)$
$=\displaystyle \lim_{n \rightarrow \infty} n\left({\big((1+\frac 1 n)^m-1\big)+\big((1+\frac 2 n)^m-1\big)+\cdots+\big((1+\frac k n)^m}-1 \big)\right)$ $=\displaystyle \lim_{n \rightarrow \infty} \left({ \frac{\big((1+\frac 1 n)^m-1\big)}{1/n}+2\frac{\big((1+\frac 2 n)^m-1\big)}{2/n}+\cdots+ k {\big(\frac{(1+\frac k n)^m-1 \big)} {k/n}}} \right)$
$=(1+2+3+\cdots+k)m$

$=\frac{k(k+1)}{2}m$

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