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

Solution 1:

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

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

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

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

$$ \sum_{i=1}^n(i^k)=P(n)=c+\frac{1}{k+1}n^{k+1}+\frac{1}{2}n^{k}+b_2kn^{k-1}+.... $$

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

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

Solution 2:

The comparison of this sum with a Riemann integral yields $$ \frac1{n\sqrt{n}}\sum_{i=1}^n\sqrt{i}=\int_0^1\sqrt{t}\mathrm dt+\sum_{i=1}^n\int_{(i-1)/n}^{i/n}(\sqrt{i/n}-\sqrt{t})\mathrm dt, $$ that is, $$ \frac1{n\sqrt{n}}\sum_{i=1}^n\sqrt{i}=\frac23+\frac1{n\sqrt{n}}\sum_{i=1}^n\int_0^1(\sqrt{i}-\sqrt{i-t})\mathrm dt, $$ or, $$ \sum_{i=1}^n\sqrt{i}=\frac23n\sqrt{n}+\sum_{i=1}^na_i,\qquad a_i=\int_0^1(\sqrt{i}-\sqrt{i-t})\mathrm dt. $$ Note that, when $i\to\infty$, $$ a_i=\int_0^1\frac{t}{\sqrt{i}+\sqrt{i-t}}\mathrm dt\sim\frac1{2\sqrt{i}}\int_0^1t\mathrm dt\sim\frac1{4\sqrt{i}}, $$ and that the series $\sum\limits_i\frac1{4\sqrt{i}}$ diverges, hence $$ \sum_{i=1}^na_i\sim\sum_{i=1}^n\frac1{4\sqrt{i}}\sim\int_0^n\frac{\mathrm dt}{4\sqrt{t}}=\frac12\sqrt{n}. $$ Using this equivalent in the formula above, one gets $$ \sum_{i=1}^n\sqrt{i}=\frac23n\sqrt{n}+\frac12\sqrt{n}+o\left(\sqrt{n}\right)=\frac23n\sqrt{n+\frac32}+o\left(\sqrt{n}\right). $$