Faulhaber's Formula to evaluate series

Question

$$S = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{99}}{n^{100}} = \lim_{n \to \infty} \frac{1}{n^{100}} \cdot \overbrace{\sum_{k=1}^{n} k^{99}}^{\text{I}}$$

$$I = \sum_{k=1}^{n} k^{99} = \frac{1}{99} \cdot \sum_{j=0}^{99} (-1)^j \binom{100}{j} B_j n^{100 - j}$$

But that seems AWFULLY, difficult? What can be done?

When dividing $I$ by $n^{100}$, only one term ($j=0$) of the sum survives, since the limit of the other terms when divided by $n^{100}$ is zero. — Jack D'Aurizio, Feb 07 '15 at 16:56
How can you bring the limit inside the sum? Can you show uniform convergence? — Ama, Feb 07 '15 at 17:08
Uniform convergence of what? This is just a sum (with a fixed number of terms), not a series of functions. — Jack D'Aurizio, Feb 07 '15 at 17:12
I see, it becomes: $$I = \frac{1}{99} \cdot \sum_{j=0}^{99} (-1)^j \binom{100}{j} B_j n^{-j}$$ What did you mean? I got something different? — Ama, Feb 07 '15 at 17:28
If you know that it is a Riemann sum, why don't use that fact? Even if you don't want to use this fact explicitly, for function monotonic over $[0,1]$, it is trivial to bound the error between the sum and the integral...
$$\left|\frac1n\sum_{k=1}^n f\left(\frac{k}{n}\right) - \int_0^1 f(x) dx \right| \le \frac1n |f(0)-f(1)|$$ — achille hui, Feb 07 '15 at 19:20
I knew this before, I was EXPLORING, a different method. @achillehui. — Ama, Feb 07 '15 at 19:25
Possible duplicate of Evaluate $\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+n^{k}}{n^{k+1}}$ — , Nov 09 '15 at 10:09

score 1 · Answer 1 · answered Feb 07 '15 at 18:40

1

Expanding Lucien's comment: $$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{99}}{n^{100}} = \lim_{n \to \infty}\frac1n\sum_{k=1}^n\Bigl(\frac{k}{n}\Bigr)^{99}=\int_0^1x^{99}\,dx. $$

answered Feb 07 '15 at 18:40

Julián Aguirre

76,354

score 1 · Answer 2 · answered Feb 07 '15 at 20:18

One doesn't need the full detail of Faulhaber's formula. The relevant fact is that $f_j(n)=\sum_{k=1}^n k^j$ is a polynomial in $n$ with leading coefficient $\dfrac{1}{j+1}$ and degree $j+1$. You can prove this by induction. Hence writing $f_j(n)=\dfrac{1}{j+1} n^{j+1}+O(n^j)$ gives that $\lim\limits_{n\to\infty}f_j(n)n^{-(j+1)}=\dfrac{1}{j+1}$.

score 0 · Answer 3 · 2015-02-07T20:12:22.963

0

By Faulhaber, $$S_n=\frac1{100}+o(\frac1n).$$

edited Feb 07 '15 at 20:12

answered Feb 07 '15 at 18:57

score 0 · Answer 4 · answered Feb 08 '15 at 00:07

Or via Cesaro-Stolz, the discrete version of l'Hopital: $$ \lim\frac{a_1+...+a_n}{b_1+...+b_n}=\lim \frac{a_n}{b_n} $$ under suitable assumptions, essentially that the second limit exists. Here $$ \lim \frac{1+2^{99}+...+n^{99}}{n^{100}}=\lim\frac{n^{99}}{n^{100}-(n-1)^{100}}=\frac1{100} $$

Faulhaber's Formula to evaluate series

4 Answers4

Linked

Related