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Let $m,\, k$ be fixed positive integers. Evaluate $$ \lim_{n \to \infty}\left\{% \left[\sum_{r = 1}^{k}{\left(n + r\right)^{m} \over n^{m - 1}}\right] -kn \right\}. $$

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  • $\begingroup$ just take lcm and combine each $n^m$ with each term , you'll get $\frac{mk(k+1)}{2}$ $\endgroup$
    – Mathronaut
    Commented Jan 25, 2014 at 17:28
  • $\begingroup$ can you tell your answer? I have solved it by putting 1/n=h and changing it to integral 0 to 1 and have got the answer $(2^{m+1}-m-2)/(m+1)$ $\endgroup$
    – user121418
    Commented Jan 25, 2014 at 18:04

1 Answer 1

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$\lim_{n \to \infty} (\frac{(\sum_{r=1}^{k}(n+r)^m)-kn^m}{n^{m-1}})= \lim_{n\to \infty}(\frac{\sum_{n=1}^{k}((n+r)^m-n^m)}{n^{m-1}})=\lim_{n\to \infty}\frac{\sum_{r=1}^{k}r((n+r)^{m-1}+\cdots + n^{m-1})}{n^{m-1}}=\lim_{n\to \infty}\sum_{r=1}^{k}r((1+\frac{r}{n})^{m-1}+(1+\frac{r}{n})^{m-2}+\cdots +(1+\frac{r}{n})+1)=\sum_{r=1}^{k} \frac{r((1+\frac{r}{n})^m)-1)}{(1+\frac{r}{n})-1}=\lim_{n \to \infty}\sum_{r=1}^{k}n((1+\frac{r}{n})^m-1)$ Now Apply L Hopital's rule to get limit as $rm$ of term inside the sum and then sum it to get $\frac{mk(k+1)}{2}$

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  • $\begingroup$ Why you expand binomially. Take n common at first step. $\endgroup$
    – user121418
    Commented Jan 26, 2014 at 8:17
  • $\begingroup$ why down vote? what will happen if you take n common? $\endgroup$
    – Mathronaut
    Commented Jan 26, 2014 at 21:46

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