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?

  • $\begingroup$ Andrei, the word "the" is a definite article. So when you say "$P$ is the polynomial function" you are saying that there is one specific polynomial function $P$ that always works. You probably were looking for the indefinite article "a". If that sentence was "$P$ is a polynomial function", then you are saying $P$ is polynomial without saying it has to be the same polynomial all the time. $\endgroup$ – Eric Stucky Jun 17 '13 at 7:37
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    $\begingroup$ What do you mean by a polynomial of non-integer degree? $\endgroup$ – Qiaochu Yuan Jun 17 '13 at 7:43
  • $\begingroup$ @EricStucky thank you very much for your suggestion. You can be sure I remembered that. $\endgroup$ – Andrei I Jun 17 '13 at 8:13
  • $\begingroup$ @QiaochuYuan technically it is not correctly formulated, but "polynomial of a non-integer degree" would mean a function of form $\sum_{k=0}^{n}a_k*x^{k+r_k}$, where $r_k$ is real and ${0<=r_k<1}$ . $\endgroup$ – Andrei I Jun 17 '13 at 8:31

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

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

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

We know Taylor expansion

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


$$ f(n-1)=f(n)-f'(n)+\frac{f''(n)}{2!}-\frac{f'''(n)}{3!}+.... $$

$$ f(n)-f(n)+f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-....=\sqrt{n} $$

$$ f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-...=\sqrt{n} $$

$$ f(n)-\frac{f'(n)}{2!}+\frac{f''(n)}{3!}-...=\int \sqrt{n} dn $$

$$ f(n)-\frac{f'(n)}{2!}+\frac{f''(n)}{3!}-\frac{f'''(n)}{4!}...=\frac{2}{3}n^\frac{3}{2} +c $$

$$ \frac{1}{2} ( f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-...)=\frac{1}{2}\sqrt{n} $$

$$ f(n)+ (-\frac{1}{2.2} +\frac{1}{3!})f''(n)+(\frac{1}{2.3!} -\frac{1}{4!})f'''(n)...=\frac{2}{3}n^\frac{3}{2}+\frac{1}{2}\sqrt{n}+c $$

$$ f''(n)-\frac{f'''(n)}{2!}+\frac{f^{4}(n)}{3!}-...)=\frac{d(\sqrt{n})}{dn}=\frac{1}{2\sqrt{n}} $$

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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    $\begingroup$ What is $c$? Why does it disappear? $\endgroup$ – Did Jun 17 '13 at 11:14
  • $\begingroup$ @ Did Thanks for feedback. If $k$ is positive integer then we know $c=0$ . but If $k$ is positive real number then I edited what it must be . Really I wonder how to find exact value of it in closed form. I need to work on it more. $\endgroup$ – Mathlover Jun 17 '13 at 11:34
  • $\begingroup$ The expansion of $f(n)$ you propose is valid for $c=0$. This is direct from the method I explained (but I have no idea about how yours can prove this). $\endgroup$ – Did Jun 17 '13 at 16:15
  • $\begingroup$ @Mathlover: Your equalies hold for $n \in \mathbb{N}$, so how do you justify the integration with respect to $n$? $\endgroup$ – Seirios Sep 12 '15 at 7:59
  • $\begingroup$ @Seirios $f(n)-f(n-1)=\sqrt{n}$ it is a function relation and I solved it for $n∈R$ via derivatives and integration. .Please think $n∈R$ for $f(n)-f(n-1)=\sqrt{n}$. The solution that I found after derivatives and integration is also true for $n∈N$ because $N∈R$. $\endgroup$ – Mathlover Sep 21 '15 at 12:51

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 $$ \sum_{i=1}^n\sqrt{i}=\frac23n\sqrt{n}+\frac12\sqrt{n}+o\left(\sqrt{n}\right)=\frac23n\sqrt{n+\frac32}+o\left(\sqrt{n}\right). $$


(Edited by user @Did on 2017, Jan 27) As explained in the comments for 3.5 years now, the correction $n+\frac54$ suggested in this answer is incorrect. For all purposes, this correction should read $n+\frac32$, see my answer for some explanations.

Since you haven't received a response yet, I can address the first bit of the question with a decent approximation,

$$\sum\limits_{i=1}^{n} \sqrt{i} \approx \frac{2}{3}n \sqrt{\frac{5}{4} + n}$$

This approximation isn't surprising, considering the fact that

$$\int\limits_{1}^{n} \sqrt{t} dt = \frac{2}{3} \left( n^{3/2} - 1\right).$$

The data looks like,

$$\newcommand\T{\Rule{0pt}{1em}{.3em}} \begin{array}{|l|l|} \hline {\rm n} & \sum\limits_{i=1}^n\sqrt {i} & \tfrac23n\sqrt{\tfrac54+n} \T \\\hline 1 \T & 1 & 1 \\\hline 5 \T & 8.382 & 8.333 \\\hline 10 \T & 22.468 & 22.361 \\\hline 25 \T & 85.634 & 85.391 \\\hline 50 \T & 239.04 & 238.63 \\\hline 100 \T & 671.4 & 670.82 \\\hline 200 \T & 1892.5 & 1891.5 \\\hline 500 \T & 7464.5 & 7462.9 \\\hline 10000 \T & 666716 & 666708 \\\hline \end{array}$$

  • $\begingroup$ Of course (and the OP seems to be fully aware of this) the sum of the first $n$ powers $i^k$ where $k$ is a constant integer is a polynomial in $n$, of degree $k+1$; see e.g. en.wikipedia.org/wiki/Faulhaber%27s_formula $\endgroup$ – Jeppe Stig Nielsen Jun 17 '13 at 8:37
  • $\begingroup$ Why $\frac54$? $ $ $\endgroup$ – Did Jun 17 '13 at 8:51
  • $\begingroup$ @Did That was just a quick approximation i came up with. You could probably come up with a better one by fiddling with that constant. I figured that we would have $n=1 \implies \text{approx}=1$ in the approximation for a better approximation for small values. So, I solved for $c$ in the following equation, $$1 = \frac{2}{3} (1) \sqrt{1 + c}$$ $\endgroup$ – Gamma Function Jun 17 '13 at 8:55
  • $\begingroup$ Unless I am mistaken $\frac32$ works better, see my answer. $\endgroup$ – Did Jun 17 '13 at 9:09
  • $\begingroup$ @Did That looks substantially better for $n$ greater than 15 or so. $\endgroup$ – Gamma Function Jun 17 '13 at 9:29

Using the Euler-Maclaurin Sum Formula, we get the asymptotic expansion for $\displaystyle\sum_{i=1}^n\sqrt{i}$ : $$ \frac23n^{3/2}+\frac12n^{1/2}+\zeta\left(-\frac12\right)+\frac1{24}n^{-1/2}-\frac1{1920}n^{-5/2}+\frac1{9216}n^{-9/2}+O\left(n^{-13/2}\right) $$ where $\zeta\left(-\frac12\right)=-\frac{1}{4\pi}\zeta\left(\frac32\right)\doteq-0.20788622497735456602$.

As described in this answer, for $\mathrm{Re}(z)\gt-1$, $$ \zeta(z)=\lim_{n\to\infty}\left(\sum_{k=1}^nk^{-z}\;-\;\left(\frac{1}{1-z}n^{1-z}+\frac12n^{-z}\right)\right) $$ The formula above agrees for $\mathrm{Re}(z)\gt1$ and the limit is analytic.

Letting $z=-\frac12$, yields that the constant in the Euler-Maclaurin Sum Formula is $\zeta\left(-\frac12\right)$.

  • $\begingroup$ thanks for the constant. (+1) Could you please add detail how you got the constant. $\endgroup$ – Mathlover Jun 17 '13 at 13:00
  • $\begingroup$ @Mathlover: that is described near the beginning of this answer. $\endgroup$ – robjohn Jun 17 '13 at 13:08
  • $\begingroup$ @Mathlover: I have added the relevant part to this post. $\endgroup$ – robjohn Jun 17 '13 at 13:20

Entering the query into WolframAlpha (query sum i=1 to n of sqrt(i)) gives two answers:

  1. By the definition of generalized harmonic numbers, the sum is the harmonic number $H_{n,{-\frac{1}{2}}}$
  2. $-\zeta\left(-\frac{1}{2}, n+1\right)-\frac{\zeta\left(\frac{3}{2}\right)}{4 \pi}$ where $\zeta$ is the Hurwitz zeta function

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