# Determine the limit for $a\geq-1$

Let $s_n(a)=1^a+2^a+\cdots+n^a$ where $a$ is real. Determine:

$$\lim_{n\to\infty}\frac{s_n(a+1)}{ns_n(a)}$$

for $a\geq-1$.

I can show that it converges. I can also find the limit for particular case when $a=0$. I think I could find a formula for natural numbers and prove it by induction. But since $a$ is real, I am stuck. Any tips of how should I proceed? Thanks!

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 Hint: this sum is approximated by the integral. – tetori Jan 22 at 14:50 you mean by $\int_1^{\infty}\frac{s_x(a+1)}{xs_x(a)}dx$ ? but it gets bloody messy, doesn't it? – Sarunas Jan 22 at 15:03 Does the limit exist when $a=-1$? – Ron Gordon Jan 22 at 15:04 it's $0$. but how could that help me? – Sarunas Jan 22 at 15:06 @Sarunas No. Just approximate the sum $s_n$. – tetori Jan 22 at 15:10

for $a>-1$,
$\lim\limits_{n\to \infty}\dfrac{\sum\limits_{k\leq n} k^{a+1}}{n\sum\limits_{k\leq n} k^a} =\lim\limits_{n\to \infty}\dfrac{\dfrac{1}{n}\sum\limits_{k\leq n} (\dfrac{k}{n})^{a+1}}{\dfrac{1}{n}\sum\limits_{k\leq n} (\dfrac{k}{n})^a} =\dfrac{\int_0^1 x^{a+1} \mathscr{d}x}{\int_0^1 x^a \mathscr{d}x} =\dfrac{a+1}{a+2}$.

for $a=-1$,
$\lim\limits_{n\to \infty}\dfrac{\sum\limits_{k\leq n} k^{a+1}}{n\sum\limits_{k\leq n} k^a} =\lim\limits_{n\to \infty}\dfrac{n}{n\sum\limits_{k\leq n} k^{-1}}=0$

So, for $a\geq 1$, $\lim\limits_{n\to\infty}\frac{s_n(a+1)}{ns_n(a)}=\dfrac{a+1}{a+2}$

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 just as tetori suggested in the comments. thanks! – Sarunas Jan 22 at 15:26

From this, $$s_m(a)=O\left(\frac{m^{a+1}}{a+1}\right)$$ (This is also evident from the examples here) for $a\ge 0$

So, $$\lim_{n\to\infty}\frac{s_n(a+1)}{ns_n(a)}$$ $$=\lim_{n\to\infty}\frac{\frac{n^{a+2}}{a+2}+\text{ terms in the lower powers of n}}{n\{\frac{n^{a+1}}{a+1}+\text{ terms in the lower powers of n}\}}$$ $$=\frac{a+1}{a+2}$$

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