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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?

Martin Sleziak
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V G
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  • 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. – Eric Stucky Jun 17 '13 at 07:37
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    What do you mean by a polynomial of non-integer degree? – Qiaochu Yuan Jun 17 '13 at 07:43
  • @EricStucky thank you very much for your suggestion. You can be sure I remembered that. – V G Jun 17 '13 at 08:13
  • @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 ${0<=r_k<1}$ . – V G Jun 17 '13 at 08:31
  • See also: http://math.stackexchange.com/questions/1431489/estimating-sum-n-1-2 and http://math.stackexchange.com/questions/79115/limit-lim-limits-n-rightarrow-infty-left2-sqrt-n-sum-limits-k-1n-frac – Martin Sleziak Sep 12 '15 at 07:44
  • $\displaystyle\sum_{k=1}^n\sqrt{k},\sim\sqrt{n}\left(\frac{2\unicode{x200a}n}{3}+\frac{1}{2}+\frac{1}{24\unicode{x200a}n}-\frac{1}{1920\unicode{x200a}n^3}+\frac{1}{9216\unicode{x200a}n^5}-\frac{11}{163840\unicode{x200a}n^7}+\mathcal O\left(n^{-9}\right)\right)-\tfrac{1}{4\unicode{x200a}\pi}\unicode{x200a}\zeta!\left(\tfrac{3}{2}\right)$ – Vladimir Reshetnikov Jul 07 '21 at 19:30

5 Answers5

19

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

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

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

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

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

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

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

Mathlover
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    What is $c$? Why does it disappear? – Did Jun 17 '13 at 11:14
  • @ 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. – Mathlover Jun 17 '13 at 11:34
  • 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). – Did Jun 17 '13 at 16:15
  • @Mathlover: Your equalies hold for $n \in \mathbb{N}$, so how do you justify the integration with respect to $n$? – Seirios Sep 12 '15 at 07:59
  • @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 $N∈R$. – Mathlover Sep 21 '15 at 12:51
  • @Mathlover, can you refer to a possible textbook where I can find this proof? Thank you – Amir Sagiv May 01 '16 at 11:36
18

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

Did
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(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,

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

Did
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Gamma Function
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  • 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. http://en.wikipedia.org/wiki/Faulhaber%27s_formula – Jeppe Stig Nielsen Jun 17 '13 at 08:37
  • Why $\frac54$? $ $ – Did Jun 17 '13 at 08:51
  • @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,

    $$1 = \frac{2}{3} (1) \sqrt{1 + c}$$

     – Gamma Function Jun 17 '13 at 08:55
  • Unless I am mistaken $\frac32$ works better, see my answer. – Did Jun 17 '13 at 09:09
  • @Did That looks substantially better for $n$ greater than 15 or so. – Gamma Function Jun 17 '13 at 09:29
  • And better for which $n$? – Did Jun 17 '13 at 09:40
  • @Did Your suggestion to use $c=\frac{3}{2}$ looks to be more accurate for large $n$. – Gamma Function Jun 17 '13 at 10:13
  • @JacobMayle a +1 for the intuition/approximation, Your answer was enough for my problem, but it seems that Mathlover's answer is the exact response for my problem (no approximation). – V G Jun 17 '13 at 10:20
  • Jacob: Sorry to insist but for which $n$? I get the impression that "for large $n$" might in fact mean "for $n\geqslant n_0$" for some $n_0$ quite small... – Did Jun 17 '13 at 11:17
  • @AndreiI "(no approximation)"... Really? Thus you use the whole series? Hence you know all its coefficients and you can sum their series?? I truly wonder how one does that in practice. – Did Jun 17 '13 at 11:19
  • @Did I didn't do the math out, but as I said in one of my previous comments, I suspect that its somewhere around 15. – Gamma Function Jun 17 '13 at 12:19
  • I just did the math, and your approximation is better for $n \geq 14$. Using $c=\frac{5}{4}$ is better for $n \leq 13$. – Gamma Function Jun 17 '13 at 12:47
  • Jacob: Thanks a bunch. – Did Jun 17 '13 at 16:17
  • @Did with "no approximation" I meant the exact computation of any coefficient. On the other side, I do not need any coefficients, as I need only the grade of the polynomial. – V G Jul 01 '13 at 16:32
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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 $\zeta\left(-\frac12\right)$.

Community
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robjohn
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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
Gerli
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