Power Sums Limits: Evaluating $\lim_{n\to \infty}$

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If $n$ and $k$ are positive integers, let $S_k(n)$ be the sum of $k$-th powers of the first $n$ natural numbers, i.e., $$S_k(n) = 1^k + 2^k + \cdots + n^k$$ Evaluate the limits $$\lim_{n\to \infty} \frac{S_k(n)}{n^k}$$ and $$\lim_{n\to\infty} \left(\frac{S_k(n)}{n^k} - \frac{n}{k+1}\right)$$

 

[anuttarasammyak]

Spoiler

$$\frac{S_1(n)}{n}=\frac{n(n+1)}{2n} \rightarrow +\infty$$
$$\frac{S_2(n)}{n^2}=\frac{n(n+1)(2n+1)}{6n^2} \rightarrow +\infty$$
$$\frac{S_3(n)}{n^3}=\frac{n^2(n+1)^2}{4n^3} \rightarrow +\infty$$
$$\frac{S_4(n)}{n^4}=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^4} \rightarrow +\infty$$
$$\frac{S_5(n)}{n^5}=\frac{n^2(n+1)^2(2n^2+2n-1)}{12n^5} \rightarrow +\infty$$
In general
$$\frac{S_k(n)}{n^k}=1+\frac{1}{k+1}\sum_{j=0}^k \ _{k+1}C_j B_j n^{1-j} =\frac{n}{k+1}+\frac{1}{2}+ o(1/n) \rightarrow +\infty$$
where $B_j$ are Bernoulli numbers.
$$\frac{S_k(n)}{n^k}-\frac{n}{k+1} \rightarrow \frac{1}{2}$$

 

[pasmith]

Spoiler

Easier is to recognise $1^k + \dots + n^k$ as the upper Darboux sum for $\int_0^n x^k\,dx$ with respect to the partition $\{0, 1, \dots, n\}$. It follows that $$1^k + \dots + n^k \geq \int_0^n x^k\,dx = \frac{n^{k+1}}{k+1}.$$ Hence $$\frac{S_k(n)}{n^k} \geq \frac{n}{k+1} \to \infty.$$

 

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Sorry, there was a term missing in the limit in the OP. I've made an edit to include the original limit.