Evaluate Limit of Faulhaber Formula

Question

Prove or evaluate that: $$ \lim_{ n \to \infty }\frac{S_{k}(n)}{n^{k+1}}=\frac{1}{k+1}$$ Where $$S_{n}^k=\sum^{n}_{m=0}m^k$$

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So I have noticed that it can be done by proving $S_k(n)$ is a polynomial of degree $n+1$ while the leading coefficient is $\frac{1}{k+1}$, but how do I expand it? Or, if this method is not recommended, can you show me how to prove it?

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And here is my incomplete attempt:

$$\sum^n_{m=0}m^k=\sum^{n}_{m=0}{(n-m)^k}=\sum^{n}_{m=0}(n-(n-m))^k=\sum^{n}_{m=0}\sum^{k}_{i=0}\binom{k}{i}n^{k-i}(m-n)^i$$ $$=\sum^{n}_{m=0} \sum^{k}_{i=0} \sum^{i}_{j=0}\binom{k}{i}\binom{i}{j}n^{k-i}m^{j}{(-n)}^{i-j}$$

Edit: Since the sequence in the denominator is strictly monotone and tending to infinity, we can use stolz theorem.

Using stolz theorem, we get \begin{align}\lim_{n\to \infty}\frac{S_k(n)}{n^{k+1}}=\\\lim_{n\to \infty}\frac{S_k({n+1})-S_k(n)}{(n+1)^{k+1}-n^{k+1}}=\\\lim_{n\to \infty}\frac{(n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\\ \lim_{n\to\infty}\frac{\sum_{i=0}^{k}\binom{k}{i}n^i}{\sum_{i=0}^{k}{\binom{k+1}{i}}n^i}=\\ \lim_{n\to\infty}\frac{n^k+\cdots}{(k+1)n^k+\cdots}=\\ \frac{1}{k+1} \\\end{align}

Answer 1

You could go a bit further using generalized harmonic numbers $$S_{n}^k=\sum^{n}_{m=0}m^k=1+H_n^{(-k)}$$ Expanding for large $n$ $$S_{n}^k= (1+\zeta (-k))+n^k \left(\frac{n}{k+1}+\frac{1} {2}+\frac{k}{12 n}+O\left(\frac{1}{n^3}\right)\right)$$

$$\frac{S_{k}(n)}{n^{k+1}}=\frac{1}{k+1}+\frac 1{2n}+\frac k{12\,n^2}+O\left(\frac{1}{n^4}\right)$$

Answer 2

Here's the start of an answer I made to a question like yours about 3 years ago. The question was closed and deleted, so I am reposting this.

If $f_k(n) = \sum_{i=1}^n i^k$ ($f_0(n) = n$) then

$\begin{array}\\ f_k(n+1) &= \sum_{i=1}^{n+1} i^k\\ &= \sum_{i=0}^{n} (i+1)^k\\ &=1+ \sum_{i=1}^{n} (i+1)^k\\ &=1+ \sum_{i=1}^{n} \sum_{j=0}^k \binom{k}{j}i^j\\ &=1+ \sum_{j=0}^k \binom{k}{j}\sum_{i=1}^{n}i^j\\ &=1+ \sum_{j=0}^k \binom{k}{j}f_j(n)\\ \text{so}\\ (n+1)^k &=f_{k}(n+1)-f_k(n)\\ &=1+ \sum_{j=0}^k \binom{k}{j}f_j(n)-f_k(n)\\ &=1+ \sum_{j=0}^{k-1} \binom{k}{j}f_j(n)\\ \end{array} $

or $kf_{k-1}(n) =(n+1)^k-1-\sum_{j=0}^{k-2} \binom{k}{j}f_j(n) $

or $(k+1)f_{k}(n) =(n+1)^{k+1}-1-\sum_{j=0}^{k-1} \binom{k+1}{j}f_j(n) $

Examples:

If $k=1$ then

$\begin{array}\\ 2f_{1}(n) &=(n+1)^{2}-1-f_0(n)\\ &=n^2+2n+1-1-n\\ &=n^2+n\\ \end{array} $

If $k=2$ then

$\begin{array}\\ 3f_{2}(n) &=(n+1)^{3}-1-f_0(n)-3f_1(n)\\ &=n^3+3n^2+3n+1-1-n-3(n^2+n)/2\\ &=\dfrac{2n^3+6n^2+6n+2-2-2n-3(n^2+n)}{2}\\ &=\dfrac{2n^3+3n^2+n}{2}\\ &=\dfrac{n(2n+1)(n+1)}{2}\\ \end{array} $

Since $(k+1)f_{k}(n) =(n+1)^{k+1}-1-\sum_{j=0}^{k-1} \binom{k+1}{j}f_j(n) $,

$f_k(n) \lt \dfrac{(n+1)^{k+1}}{k+1} $.

Also, $f_k(n)$ is a polynomial of degree $k+1$. with constant term $0$.

(I have more details on getting the coefficients of the polynomial, but I'll stop here.)

Answer 3

Using Cesaro-Stolz, as you pointed out, we just need to evaluate $\lim_{n\to\infty} \tfrac{1}{n+1-n\big(\tfrac{n}{n+1}\big)^k}$. So it suffices to show that $\lim_ {n\to\infty} n\big(1-(\tfrac{n}{n+1})^k\big) = k$.

$\big(\tfrac{n}{n+1}\big)^k=\big(1-\tfrac{1}{n+1}\big)^k=\sum_{j=0}^k {k\choose j}1^j(-1)^{k-j}\tfrac{1}{(n+1)^{k-j}}=1-\tfrac{k}{n+1}+\frac{1}{(n+1)^2}A$ for some expression $A$ that is bounded as $n\to\infty$.

So, $\lim_{n\to\infty} n\big(1-(\tfrac{n}{n+1})^k\big)=\lim_{n\to\infty}\big(k\tfrac{n}{n+1}-A\tfrac{n}{(n+1)^2}\big)=k$.

Answer 4

$\sum_{i=1}^n i^k=\sum_{j=1}^{k+1} c_jn^j$

Since we are dividing the $n^{k+1}$, the only non-zero term is the coefficient of $n^{k+1}$.

$(n+1)^k= \sum_{j=1}^{k+1} c_j[(n+1)^j-n^j]=\sum_{j=1}^{k+1}\sum_{q=0}^{j-1}c_j\binom{j}{q}n^q$

Coefficient for $n^k$ term has to be th same on both side, i.e. 1.

$c_{k+1}\binom{k+1}{k}=1 \implies c_{k+1}\frac{(k+1)!}{k!1!}\implies c_{k+1}=\frac{1}{k+1}$

Answer 5

By far the simplest approach is to use Riemann sums $$ \frac{1}{n^{k+1}} \sum_{m=0}^n m^k = \frac{1}{n} \sum_{m=0}^n \left(\frac{m}{n}\right)^k \xrightarrow[n\to\infty]{} \int_0^1 x^k \text dx = \frac{1}{k+1}. $$