What is n∑i=1√i ?
Also I noticed that n∑i=1ik=P(n)
where k is a natural number and P is a polynomial of degree k+1. Does that also hold for any real positive number? How could one prove it?
Answer
n∑i=1√i=f(n)
n−1∑i=1√i=f(n−1)
f(n)−f(n−1)=√n
We know Taylor expansion
f(x+h)=f(x)+hf′(x)+h2f″
Thus
f(n-1)=f(n)-f'(n)+\frac{f''(n)}{2!}-\frac{f'''(n)}{3!}+....
f(n)-f(n)+f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-....=\sqrt{n}
f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-...=\sqrt{n}
f(n)-\frac{f'(n)}{2!}+\frac{f''(n)}{3!}-...=\int \sqrt{n} dn
f(n)-\frac{f'(n)}{2!}+\frac{f''(n)}{3!}-\frac{f'''(n)}{4!}...=\frac{2}{3}n^\frac{3}{2} +c
\frac{1}{2} ( f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-...)=\frac{1}{2}\sqrt{n}
f(n)+ (-\frac{1}{2.2} +\frac{1}{3!})f''(n)+(\frac{1}{2.3!} -\frac{1}{4!})f'''(n)...=\frac{2}{3}n^\frac{3}{2}+\frac{1}{2}\sqrt{n}+c
f''(n)-\frac{f'''(n)}{2!}+\frac{f^{4}(n)}{3!}-...)=\frac{d(\sqrt{n})}{dn}=\frac{1}{2\sqrt{n}}
If you continue in that way to cancel f^{r}(n) terms step by step, you will get
f(n)=c+\frac{2}{3}n^\frac{3}{2}+\frac{1}{2}\sqrt{n}+a_2\frac{1}{\sqrt{n}}+a_3\frac{1}{n\sqrt{n}}+a_4\frac{1}{n^2\sqrt{n}}+....
You can find a_n constants by Bernoulli numbers, please see Euler-Maclaurin formula. I just wanted to show the method. http://planetmath.org/eulermaclaurinsummationformula
You can also apply the same method for \sum_{i=1}^n(i^k)=P(n), k is any real number .
you can get
\sum_{i=1}^n(i^k)=P(n)=c+\frac{1}{k+1}n^{k+1}+\frac{1}{2}n^{k}+b_2kn^{k-1}+....
P(1)=1=c+\frac{1}{k+1}+\frac{1}{2}+b_2k+....
=c=1-\frac{1}{k+1}-\frac{1}{2}-b_2k+....
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