4 Answers4

15

Your sum can be interpreted as a Riemann sum:

$$\sum_{k=1}^n \frac{1}{\sqrt{kn}} = \frac1n \sum_{k=1}^n \sqrt{\frac{n}{k}}. $$

Let $f(x) = 1/\sqrt{x}$ and let $x_k = k/n$. Then $$\sum_{k=1}^n \frac{1}{\sqrt{kn}} = \frac1n \sum_{k=1}^n \sqrt{\frac{n}{k}} = \frac1n \sum_{k=1}^n f(x_k) \to \int_0^1 f(x)\,dx $$ as $n \to \infty$.

(Since the integral is improper, a little care is needed to justify the last step.)

mrf
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  • How we can justify the last step? – Pedro Sep 02 '13 at 13:07
  • So we need to prove that $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nf(x_k)=\lim_{\varepsilon\to 0^+}\left(\lim_{n\to\infty}\frac{(1-\varepsilon)}{n}\sum_{k=1}^nf(x_k)\right)$$ Right? – Pedro Sep 02 '13 at 13:33
  • A good idea. The caution at the end is a good one ... Riemann sums for a convergent improper integral may or may not go to the value of the integral. – GEdgar Sep 02 '13 at 13:42
  • @GEdgar At first, we don't konw if $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nf(x_k)$$ converges (thus limit properties are not valid). So, how we can to prove the equality above? – Pedro Sep 02 '13 at 13:56
  • This method is well known on physics community. It goes like this: $\xi \equiv k/n$. $\Delta\xi = 1/n \Longrightarrow n\Delta\xi = 1$. Then, $\displaystyle{\sum_{1}^{n}\left(kn\right)^{-1/2} = \sum_{1}^{n}\left(n\xi n\right)^{-1/2}n,\Delta\xi \to \int_{1}^{\infty}\xi^{-1/2},{\rm d}\xi}$ ( when $n \to \infty$ ) which is the quite fine @mrf answer. – Felix Marin Sep 02 '13 at 16:16
  • Actually, the sum you give in your answer, if converted to integrals of step functions, converges to $f$ by Monotone Convergence. – robjohn Sep 04 '13 at 15:55
  • This is a correction to my comment (above): $\displaystyle{\sum_{1}^{n}\left(kn\right)^{-1/2} = \sum_{1}^{n}\left(n\xi n\right)^{-1/2}n,\Delta\xi \to \int_{1/n}^{1}\xi^{-1/2},{\rm d}\xi = 2 - 1/n}$ – Felix Marin Sep 05 '13 at 19:49
10

Notice for large $n$, we expect $\displaystyle \sum_{k=1}^n \frac{1}{\sqrt{k}} \text{ behave like }\int_1^n \frac{dx}{\sqrt{x}} \sim 2\sqrt{n}$. This suggests $$\frac{1}{\sqrt{k}} \sim \int_{k-1/2}^{k+1/2} \frac{dx}{\sqrt{x}} \sim 2\left( \sqrt{k+\frac12} - \sqrt{k-\frac12}\right)$$ and the terms $\displaystyle \frac{1}{\sqrt{kn}}$ in the summands is close to something "telescopable". To make this idea concrete, we observe: $$\begin{align} \sum_{k=1}^n\frac{1}{\sqrt{kn}} \ge & \sum_{k=1}^n \frac{2}{\sqrt{n}(\sqrt{k+1}+\sqrt{k})} = \frac{2}{\sqrt{n}} \sum_{k=1}^n(\sqrt{k+1}-\sqrt{k}) = 2 \Big(\sqrt{1+\frac{1}{n}} - \frac{1}{\sqrt{n}}\Big)\\ \sum_{k=1}^n\frac{1}{\sqrt{kn}} \le & \sum_{k=1}^n \frac{2}{\sqrt{n}(\sqrt{k}+\sqrt{k-1})} = \frac{2}{\sqrt{n}}\sum_{k=1}^n(\sqrt{k}-\sqrt{k-1}) = 2 \end{align}$$

As a result, $$\left|\;\sum_{k=1}^n \frac{1}{\sqrt{kn}} - 2\;\right| \le 2 \left(1+\frac{1}{\sqrt{n}} -\sqrt{1+\frac{1}{n}}\right) < \frac{2}{\sqrt{n}} \quad\implies\quad \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}} = 2.$$

achille hui
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  • 1
    +1 Nice. An informal motivation would be to note that (for large $k$, using first order Taylor expansion) $\sqrt{k+1} \approx \sqrt{k} + 1/(2\sqrt{k})$, or $1/\sqrt{k} \approx 2(\sqrt{k+1} - \sqrt{k})$ which can be regarded as a discrete analogous to the derivative $(2\sqrt{x})' = 1/\sqrt{x}$ – leonbloy Sep 02 '13 at 15:12
7

mrf has the main idea. But since the integral is improper (as mrf notes) some care is required. Here is one way, using the Lebesgue theory...

Let $f(x) = 1/\sqrt{x}$. For fixed $n$, let $g_n$ be defined by $g_n(x) = \sqrt{n/k}$ for $(k-1)/n < x \le k/n$. Then $0 < g_n(x) \le f(x)$ and $g_n(x) \to f(x)$ on $(0,1]$. Since $f$ is Lebesgue integrable on $(0,1]$, we have by the dominated convergence theorem $\int_0^1 g_n \to \int_0^1 f$. That is: $$ \frac{1}{n}\sum_{k=1}^n\sqrt{\frac{n}{k}} \to \int_0^1 \frac{dx}{\sqrt{x}} $$

ASIDE

Monthly problem 11376 has an example of how blindly saying "Riemann sum" can lead one astray. The solution is on p. 283 of the March, 2010 issue. The problem defines $$ S_n(a) = \sum_{an \lt k \le (a+1)n}\frac{1}{\sqrt{kn-an^2}\;} $$ for real $a$ and positive integer $n$, and asks for which $a$ does $\lim_{n \to \infty} S_n(a)$ exist. Many solvers noted that $S_n(a)$ is a Riemann sum for $$ \int_a^{a+1} \frac{dx}{\sqrt{x-a}\;} = 2 $$ and then carelessly concluded that $S_n(a) \to 2$ for all $a$. But, in fact, as the published solution shows, $S_n(a)$ converges if and only if $a$ is rational.

The problem here is the case $a=0$, and fortunately $0$ is rational.

GEdgar
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  • Ha ha! I am one of those who falsely "proved" convergence. –  Sep 02 '13 at 16:00
  • As I mentioned in my comment to mrf's answer, I believe that Monotone Convergence also applies since this sequence of functions is pointwise monotone. – robjohn Sep 04 '13 at 15:58
  • @robjohn: It is not monotone along $n$, but yes it is monotone along $2^n$, say. – GEdgar Sep 04 '13 at 16:07
  • @GEdgar I would like know how to prove that $S_n(a)$ converges if and only if $a$ is rational. Is it possible you post the solution of this problem? – Pedro Sep 04 '13 at 16:37
  • Monthly problem 11376 of what journal? – Robin Oct 25 '18 at 15:26
  • "Monthly" means "American Mathematical Monthly" ... Solution by Vitali Stakhovsky is in the March 2010 issue, vol 11, no 3, p. 282. If your library subscribes to the Monthly, see it there. You may try https://www.scribd.com/document/354083365/AMM-2010-N3 and scroll down to p 94 of 101 ... but since that page is presumably illegal, it may not be around for long. – GEdgar Oct 25 '18 at 16:11
5

Think about $$\int_0^1 f(x)dx,~~~f(x)=\frac{1}{\sqrt{x}} $$

Indeed: $$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{i=1}^nf\left(a+\frac{b-a}{n}i\right)\left(\frac{b-a}{n}\right)$$

Mikasa
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  • I think your solution is valid if $$\lim_{n\to\infty}\sum_{k=1}^nf\left(\frac{i}{n}\right)\frac{b}{n}=\lim_{a\to 0^+}\left(\lim_{n\to\infty}\sum_{k=1}^nf\left(a+\frac{b-a}{n}i\right)\frac{(b-a)}{n}\right)$$ So, how to prove it? – Pedro Sep 02 '13 at 14:16
  • Nice work, Babak! +1 – amWhy Sep 03 '13 at 00:11