summation - What is $sumlimits_{i=1}^n sqrt i $?

What is $\sum\limits_{i=1}^n\sqrt i\ $?

Also I noticed that $\sum\limits_{i=1}^ni^k=P(n)$
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

$$
\sum_{i=1}^n\sqrt{i}=f(n)
$$

$$
\sum_{i=1}^{n-1}\sqrt{i}=f(n-1)

$$

$$
f(n)-f(n-1)=\sqrt{n}
\tag 1$$

We know Taylor expansion

$$
f(x+h)=f(x)+hf'(x)+\frac{h^2 f''(x)}{2!}+\frac{h^3f'''(x)}{3!}+....

$$

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}+....

$$

$$
P(1)=1=c+\frac{1}{k+1}+\frac{1}{2}+b_2k+....
$$

$$
=c=1-\frac{1}{k+1}-\frac{1}{2}-b_2k+....
$$