Will You Find the 100th Power Sum Formula?

Using Riemann sums , the given limit is

$\displaystyle\lim_{n \to \infty} \dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^{100} = \int_{0}^{1} x^{100} dx = \left(\dfrac{x^{101}}{101}\right)_{0}^{1} = \dfrac{1}{101}$ . Thus $A + B = 1 + 101 = \boxed{102}$ .

It is possible to have an idea of what a summation to the $100^{th}$ power looks like, by doing:

$\large \displaystyle 1^{101} = (1+0)^{101} = C_{0}^{101} 0^{101} + C_{1}^{101} 0^{100} + C_{2}^{101} 0^{99} + ... + C_{100}^{101} 0^{1} + C_{101}^{101}$

$\large \displaystyle 2^{101} = (1+1)^{101} = C_{0}^{101} 1^{101} + C_{1}^{101} 1^{100} + C_{2}^{101} 1^{99} + ... + C_{100}^{101} 1^{1} + C_{101}^{101}$

$\large \displaystyle 3^{101} = (1+2)^{101} = C_{0}^{101} 2^{101} + C_{1}^{101} 2^{100} + C_{2}^{101} 2^{99} + ... + C_{100}^{101} 2^{1} + C_{101}^{101}$

$\large \displaystyle \vdots$

$\large \displaystyle (1+n)^{101} = C_{0}^{101} n^{101} + C_{1}^{101} n^{100} + C_{2}^{101} n^{99} + ... + C_{100}^{101} n^{1} + C_{101}^{101}$

Summing up all the equations above, each term in the L.H.S. will cancel out with the first term on the R.H.S. of the next equation. What will remain is:

$\large \displaystyle (1+n)^{101} = C_{1}^{101} \sum_{k = 1}^n k^{100} + C_{2}^{101} \sum_{k = 1}^n k^{99} + ... + C_{100}^{101} \sum_{k = 1}^n k + C_{101}^{101} \sum_{k = 1}^n 1$

$\large \displaystyle n^{101} + O(n^{100}) = C_{1}^{101} \sum_{k = 0}^n k^{100} + C_{2}^{101} \sum_{k = 0}^n k^{99} + ... + C_{100}^{101} \sum_{k = 0}^n k + C_{101}^{101} \sum_{k = 0}^n 1$

$\large \displaystyle n^{101} + O(n^{100}) = C_{1}^{101} \sum_{k = 1}^n k^{100} + C_{2}^{101} \sum_{k = 1}^n k^{99} + ... + C_{100}^{101} \sum_{k = 1}^n k + C_{101}^{101} \sum_{k = 0}^n 1$

Since a sum from $1$ to $n$ with each term to the $a^{th}$ power has order $a+1$ :

$\large \displaystyle n^{101} + O(n^{100}) = 101 \sum_{k = 1}^n k^{100} + O(n^{100})$

$\color{#20A900} \boxed{\large \displaystyle \sum_{k = 1}^n k^{100} = \frac{n^{101}}{101} + O(n^{100}) }$

So:

$\large \displaystyle \lim _{n \to \infty} \frac{1^{100} + 2^{100} + 3^{100} + ... + n^{100}}{n^{101}}$

$\large \displaystyle = \lim _{n \to \infty} \frac{ \frac{n^{101}}{101} + O(n^{100})}{n^{101}}$

$\color{#20A900} =\boxed{\large \displaystyle \frac{1}{101} }$

So:

$\color{#3D99F6} \large \displaystyle A=1, B = 101, \boxed{\large \displaystyle A+B = 102}$

It is a pattern which i found for such problems. Every nth degree sum has its value as1/(n+1)x. The question above is clear example. Any proof in simple words will be greatly appreciated.

I used Faulhaber's Formula .

It says that the sum of the $p^{th}$ powers of the first $n$ natural numbers can be written as a $(p+1)^{th}$ degree polynomial function of $n$ .

$\displaystyle \large \sum _{k=1}^nk^p=\frac{1}{p+1}\sum _{i=0}^p\left(-1\right)^i\binom{p+1}{i}B_i n^{p+1-i}$ , where $B_i$ denotes the $i^{th}$ Bernoulli Number .

For our case, $p=100$ :

$\displaystyle \large \frac{1}{101}\sum _{i=0}^p\left(-1\right)^i\binom{101}{i}B_i n^{101-i}$

Now since $n \rightarrow \infty$ , terms with degree lesser than 101 will approach 0. ( $\frac{n^k}{n^{101}} = n^{k-101}$ will converge to 0 if $k<101$ ).

So, only the first term will survive.

$\displaystyle \large \lim_{n \to \infty} \frac{1}{101}\frac{\left(-1\right)^0\binom{101}{0}B_0 n^{101-0}}{n^{101}} = \frac{1}{101} (-1)^0 \binom{101}{0} B_0 = \frac{1}{101} \cdot 1 \cdot 1 \cdot 1 = \boxed{\frac{1}{101}}$