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The value of $$\lim_{n=\infty} \dfrac {1^{a+1}+2^{a+1}+\cdots+n^{a+1}}{n.(1^{a }+2^{a }+\cdots+n^{a })} $$

Attempt: $S = \lim_{n \rightarrow \infty} \sum_{n=0} ^\infty \dfrac {k^{a+1}} {n.( 1^{a }+2^{a }+\cdots+ n^{a} )}$

I am trying to use the integral as a limit of sum but the denominator is preventing me from applying.

Could anyone please give me a direction to move on.

Thank you for your help.

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The limit can be interpreted as the ratio of two Riemann sums. First of all we have that

$$I_k = \lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n \left(\frac{i}{n}\right)^k = \int_0^1 x^k dx = \frac{1}{k+1}$$

which gives $$S = \lim_{n\to \infty}\frac{\frac{1}{n}\sum_{i=1}^n \left(\frac{i}{n}\right)^{a+1}}{\frac{1}{n}\sum_{i=1}^n \left(\frac{i}{n}\right)^{a}} = \frac{I_{a+1}}{I_a} = \frac{a+1}{a+2}$$

where in the second step above I have used the fact that if $a_n$ and $b_n$ converges then $\lim_{n\to\infty} \frac{a_n}{b_n} = \frac{\lim_{n\to\infty} a_n}{\lim_{n\to\infty} b_n}$.

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  • $\begingroup$ Great Answer :) Thank you. I have a question though please. Shouldn't $I_{a+1}$ be equal to $\dfrac {1}{a+2}?$ $\endgroup$ – MathMan Jan 27 '15 at 18:26
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    $\begingroup$ @Wanderer You are absolutely right, I have corrected it now. Thanks! $\endgroup$ – Winther Jan 27 '15 at 19:14
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Hint: $1^a+2^a+\cdots+n^a\sim {n^{a+1}\over a+1}$.

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  • $\begingroup$ Thank you for the hint !:) $\endgroup$ – MathMan Jan 27 '15 at 18:35

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