By looking at an integral and bounding the error?

  • 3
    $\begingroup$ What have you tried so far? $\endgroup$ Sep 29, 2010 at 2:50
  • $\begingroup$ Yes I have tried. I got bounds on the sum by but they differ by order of \sqrt{n} which doesn't seem like a great estimate. $\endgroup$
    – blaklaybul
    Sep 29, 2010 at 14:39
  • $\begingroup$ @blaklaybul Compare the sum with $\int_0^n\sqrt{x}dx=2n\sqrt{n}/3$. $\endgroup$ Sep 30, 2010 at 21:25
  • $\begingroup$ @AD.: I suggest you check out the answers. What you said has already been said in the answers and blaklaybul's comment about sqrt(n) error is exactly about that, I believe! $\endgroup$
    – Aryabhata
    Oct 1, 2010 at 6:23
  • $\begingroup$ @Moron: Sorry, i was a bit hasty (btw it was a nice job you did there). What I wanted to say is that it is easy to get a feel for a sum by looking at a similar integral (which are often much easier to deal with of course). Maybe too trivial comment? $\endgroup$ Oct 1, 2010 at 9:09

7 Answers 7


This was an interesting challenge, to try and come up with an estimate for this sum in an elementary way.

The estimate I got was

$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$

for some constant $C$ which appears to be close to $-0.207$ (which I guess will be $\zeta(-1/2)$).

(By $a_{n} \sim b_{n}$ I mean $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0$)

I believe here is a completely elementary proof of that fact:

First consider the inequality for $x > 0$ and $k > 0$.

$$ \sqrt{x} \le \frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}$$

This follows easily by the arithmetic mean $\ge$ geometric mean inequality.

Thus we have that

$$\int_{k}^{k+1} \sqrt{x} \ dx \le \int_{k}^{k+1} (\frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}) \ dx$$

$$ = \frac{(k+1)^2 - k^2}{4\sqrt{k}} + \frac{\sqrt{k}}{2} = \sqrt{k} + \frac{1}{4\sqrt{k}}$$

Thus $$\sum_{k=1}^{n-1} \int_{k}^{k+1} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$


$$\int_{1}^{n} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$

and so

$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$

Now we have inequality

$$\frac{1}{2\sqrt{k}} < \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sqrt{k} -\sqrt{k-1}$$

And so, we have that

$$ \sum_{k=1}^{n-1} \frac{1}{4\sqrt{k}} < \frac{\sqrt{n-1}}{2}$$


$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \frac{\sqrt{n-1}}{2} + \sum_{k=1}^{n-1} \sqrt{k} $$

So if $$S_{n} = \sum_{k=1}^{n} \sqrt{k}$$ we have that

$$S_{n} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \sqrt{n} - \frac{\sqrt{n-1}}{2} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \frac{\sqrt{n}}{2}$$

Now let $$G_{n} = S_{n} - \frac{2}{3} n^{3/2} - \frac{\sqrt{n}}{2}$$

We have that $$G_{n} \ge -\frac{2}{3}$$

We can easily show that (using tedious but not too complicated algebra*) $$G_{n+1} < G_{n}$$ and so $G_{n}$ is a convergent sequence, as it is bounded below and monotonically decreasing.

Thus there exists a constant $C$ (the limit of $G_{n}$) such that

$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$

* For the sake of completeness, we show that $G_{n+1} < G_{n}$.

Consider $$6(G_{n}-G_{n+1}) = 4(n+1)\sqrt{n+1} + 3\sqrt{n+1} - 6\sqrt{n+1} - 4n\sqrt{n} - 3\sqrt{n}$$ $$ = \sqrt{n+1} + 4n(\sqrt{n+1} -\sqrt{n}) - 3\sqrt{n}$$

Multiplying by $\sqrt{n+1} + \sqrt{n}$ does not change the sign, so we look at

$$ \sqrt{n+1}(\sqrt{n+1} + \sqrt{n}) + 4n - 3\sqrt{n}(\sqrt{n+1} + \sqrt{n})$$ $$ = 2n+1 - 2\sqrt{n^2 + n}$$

Now $$ (2n+1)^2 = 4n^2 + 4n + 1 > 4n^2 + 4n = (2\sqrt{n^2+n}) ^2$$


$$ 2n+1 > 2\sqrt{n^2+n}$$

and so

$$G_{n} > G_{n+1}$$

  • 10
    $\begingroup$ The constant is $\zeta(-1/2)$. The Riemann zeta-function can be continued analytically by the Euler-Maclaurin summation method and doing so ties up the constant term here with the zeta-value. $\endgroup$ Sep 29, 2010 at 20:10
  • 1
    $\begingroup$ Very nice. More interesting than anticipated. $\endgroup$ Sep 29, 2010 at 23:38

In case you were wondering, like me, Moron's excellent proof adapts easily to show that

$$1 + \sqrt[3]{2} + \dots + \sqrt[3]{n} \sim \frac{3}{4}n^{4/3} + \frac{\sqrt[3]{n}}{2} + C,$$

for some constant $C.$ In this case $C = \zeta(-1/3) \approx -0.277343.$

Where, as before, $a_n \sim b_n$ means $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0.$

Similar to the previous proof, we use the AM-GM inequality to show

$$\sqrt[3]{x} \le \frac{x}{3k^{2/3}} + \frac{2k^{1/3}}{3}.$$

Summing from $k=1$ to $n-1$ and integrating we arrive at $$ \frac{3}{4}n^{4/3} - \frac{3}{4} \le \sum_{k=1}^{n-1}\sqrt[3]{n} + \frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n>1.$$ And so, $$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + n^{1/3} - \frac{3}{4} - \frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n>1.$$

Using $\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le 1+ \int_1^n x^{-2/3} dx, \qquad n>1,$ we obtain

$$\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le \frac{1}{2}n^{1/3} - \frac{1}{3}.$$

And hence $$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + \frac{1}{2}n^{1/3} - \frac{5}{12}, \qquad n>1.$$

As in the previous argument, set

$$G_n = \sum_{k=1}^{n}\sqrt[3]{n} - \frac{3}{4}n^{4/3} - \frac{1}{2}n^{1/3}.$$

Then $G_n \ge -5/12,$ and we can show $G_n – G_{n+1} > 0$ by showing that $$\frac{(n+1)^{4/3} – n^{4/3}}{(n+1)^{1/3} + n^{1/3}} > \frac{2}{3}.$$

So, as before, $G_n$ is convergent, since it is bounded below and monotonically decreasing.

I suspect this argument also adapts easily to the more general case $\sum_{k=1}^{n}\sqrt[r]{n},$ for $r \in \mathbb{N},$ where I'm guessing, we'll find $$1 + \sqrt[r]{2} + \dots + \sqrt[r]{n} \sim \frac{r}{r+1}n^{(r+1)/r} + \frac{\sqrt[r]{n}}{2} + C,$$ where $C = \zeta(-1/r).$ However, I confess to not having checked the details of this further generalisation.


The "sophisticated" way to do this is using the Euler-Maclaurin formula - look under the heading "Asymptotic expansion of sums". You'll want to start your sum at 1, not at 0, to avoid dividing by zero.


The standard approach should be along the lines of $\sum_{i\le n}\sqrt i=\sqrt n\sum_{i\le n}\sqrt{\frac{i}{n}}$, so $\frac1{n\sqrt n}\sum_{i\le n}\sqrt i=\sum_{i\le n}\sqrt{\frac{i}{n}}\frac1n\to\int_0^1\sqrt x{} dx=\frac23$, or $\sum_{i\le n}\sqrt i\sim\frac23 n\sqrt n$. What techniques do you know to estimate the error between integrals and their Riemann sums?

  • $\begingroup$ This is a problem I found on a blackboard in the math department. I saw someone trying to approach this using this trapezoidal approximations. $\endgroup$
    – blaklaybul
    Sep 29, 2010 at 14:37

Since square root is a monotonically increasing function, that summation will be between the integral of sqrt(i) from 0 to n and the integral of sqrt(i) from 1 to n-1.


You will find some really nice approximations at MathKB.

PS: You'll also find a couple of exact formulas there, though not written with elementary functions of course.


By the Euler-Maclaurin summation formula, we have

\begin{align}\sum_{k=1}^n\sqrt k&=\frac23n^{3/2}+\frac12n^{1/2}+\zeta(-1/2)+\sum_{k=1}^\infty\frac{B_{2k}\Gamma(\frac32)}{(2k)!\Gamma(\frac52-2k)}n^{\frac32-2k}\\&=\frac23n^{3/2}+\frac12n^{1/2}+\zeta(-1/2)+\frac1{24}n^{-1/2}+\frac1{1920}n^{-3/2}+\mathcal O(n^{-7/2})\end{align}

Seeing as the remainder term in the expansion goes to zero for $n\ge1$ and the collected constants add up to the zeta function.

A graph of the terms as far as expanded is shown below:

enter image description here

More generally, we have, for $s\ne-1$ and large enough $n$,


As implimented in this graph.

  • $\begingroup$ If you use the asymptotic form of the Bernoulli numbers you can estimate for what $k$ you'll get the minimum terms and also estimate the value of that minimum term. Truncating at that point is optimal, the error is then bounded by that minimum term and that is exponentially small as a function of $n$. You can do even better by taking the divergent tail and resumming that using Borel summation. This yields an integral that you can approximate using the saddle point method, this will yield another asymptotic series. $\endgroup$ Sep 8, 2017 at 23:22
  • $\begingroup$ @CountIblis Oh, fun, I think I'll try Borel summing later. $\endgroup$ Sep 8, 2017 at 23:24

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