How to prove the following equality involving sums

Question

I want to prove the following :

$\lim\limits_{n\to\infty}\frac{1}{n^a}$$\sum\limits_{k=1}^{n}k^{a-1}$=$\frac{1}{a}$ , $\forall a \geq 1 , a \in N$

Now it is obvious that this holds for $a=1,2,3 ...$ but how will one obtain the sum in compact form for $a=t+1$ ?

Answer 1

Using Riemann sum theorem $\lim_{n\to \infty}\dfrac{1}{n^a}\sum_{k=1}^{n}k^{a-1} = \lim_{n\to \infty}\dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^{a-1} = \int_0^1x^{a-1}dx = \dfrac{x^a}{a}|_0^1 = \dfrac{1}{a} $

Answer 2

It is easier to work in a direct way.

$$S_d(n):=\sum_{k=0}^n k^d$$

is a polynomial of degree $d+1$, because $S_d(n)-S_d(n-1)=n^d$ is a polynomial of degree $d$. If the leading term is $s_dn^{d+1}$, then using the binomial theorem,

$$s_dn^{d+1}-s_d(n-1)^{d+1}=(d+1)s_dn^d+\cdots$$ terms of lower degree.

This must match $n^d$ and

$$s_d=\frac1{d+1}.$$

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For fun, the next coefficient:

$$s_dn^{d+1}+s_{d-1}s^d-s_d(n-1)^{d+1}-s_{d-1}(n-1)^d\\ =(d+1)s_dn^d-\frac{(d+1)d}2n^{d-1}+ds_{d-1}n^{d-1}+\cdots$$

gives after simplification (matching $n^d+0n^{d-1}$),

$$s_d=\frac1{d+1},\\s_{d-1}=\frac12.$$

This matches the Faulhaber formula.