How to prove that

$$\lim_{n\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$

using only basic limit operations and theorems?


closed as off-topic by José Carlos Santos, Nosrati, Micah, Connor Harris, Gibbs Oct 1 '18 at 17:08

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  • 8
    $\begingroup$ Have you made any attempts to solve this problem? If you include what you've tried in the body of your question, we can help you better. $\endgroup$ – Carl Schildkraut Oct 1 '18 at 15:20
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    $\begingroup$ Besides, there arrives a question, what is a basic limit operation, and what isn't. $\endgroup$ – Jakobian Oct 1 '18 at 15:25
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    $\begingroup$ @Jakobian. Basic arithmetic operations with limits, Bernoulli's inequality, Second Remarkable Limit $\endgroup$ – Samson Oct 1 '18 at 15:34
  • $\begingroup$ Are you allowed to use the fact that the series $\sum_{i\geq 1}\frac{1}{i}$ diverges? $\endgroup$ – Boshu Oct 1 '18 at 15:35
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    $\begingroup$ @Samson you should include that in your question instead of telling me :) $\endgroup$ – Jakobian Oct 1 '18 at 15:36

You have $$\begin{align} \frac{1}{n}\sum_{k=1}^n \sqrt{k} &\geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{k} \geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{\frac{n}{2}} \\ &= \frac{n-\lfloor n/2\rfloor+1}{n}\cdot \sqrt{\frac{n}{2}} \geq \frac{n}{2n}\cdot \sqrt{\frac{n}{2}} \\ &= \frac{\sqrt{n}}{2\sqrt{2}} \xrightarrow[n\to\infty]{} \infty \end{align}$$ where the only "trick" lies in the first inequality: dropping some (positive) terms from the sum actually helps to get an easy lower bound on the sum.

  • $\begingroup$ +1. very nice answer! $\endgroup$ – Taladris Oct 1 '18 at 15:43
  • $\begingroup$ @Taladris Thanks! $\endgroup$ – Clement C. Oct 1 '18 at 15:44

Using the Stolz-Cesaro Theorem, $$\lim_{n\to\infty}\frac{1+\sqrt{2}+...+\sqrt{n}}{n}=\lim_{n\to\infty} \sqrt{n}=\infty$$

  • $\begingroup$ +1. The straightforward way. $\endgroup$ – Felix Marin Oct 4 '18 at 1:23

WEll, note the inequality $\sqrt{i} \ge \frac{i}{n} \times \sqrt{n}$ holds for each $i \le n$ where both $i$ and $n$ are nonnegative. [Make sure you can see it]

Thus, the following string of inequalities hold:

$$\frac{1}{n} \sum_{i=1}^ n \sqrt{i} \ \ge \ \frac{1}{n} \sum_{i=1}^ n \frac{i}{n} \times \sqrt{n} \ = \ \frac{\sqrt{n}}{n^2}\sum_{i=1}^n i \ = \ \frac{\sqrt{n}}{n^2} \theta(n^2), $$ which as $n$ goes to infinity, gives the claimed bound.


Creative telescoping does the job nicely. We have $$(n+1)\sqrt{n+1}-n\sqrt{n} = \sqrt{n+1}+\frac{n}{\sqrt{n}+\sqrt{n+1}} \leq \frac{3}{2}\sqrt{n+1}$$ hence $$\sum_{k=1}^{n}\sqrt{k}\geq \frac{2}{3}\sum_{k=1}^{n}\left[n\sqrt{n}-(n-1)\sqrt{n-1}\right] = \frac{2}{3}n\sqrt{n}.$$


Note: I am not sure what count as a "basic operation" on limits.

Method $1$ (inspired by the classic proof of Integral Test for Series):

Let $S_n=\sum_{k=1}^n \sqrt{k}$. The function $f(x)=\sqrt{x}$ is increasing so, for any integer $k$, we have

$$\sqrt{k}\leqslant \sqrt{x} \leqslant \sqrt{k+1}$$

for $x\in[k,k+1]$. Integrating the inequality on $[k,k+1]$, we obtain

$$\sqrt{k}\leqslant \int_{k}^{k+1}\sqrt{x}\; dx\leqslant \sqrt{k+1}$$

Summing from $k=1$ to $n$, we have

$$S_n \leqslant \int_{1}^{n+1}\sqrt{x} \leqslant S_{n+1}-1$$

The middle integral is $\frac{2}{3}(n+1)^{3/2}-1$, so we obtain that

$$S_n \geqslant \frac{2}{3}n^{3/2}$$

so $\frac{S_n}{n} \geqslant \frac{2}{3}n^{1/2}$. Since $\lim_{n\to \infty} n^{1/2}=\infty$, we obtain that

$$\lim_{x\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$

"Method $2$": I believe that the inequality $\frac{S_n}{n} \geqslant \frac{2}{3}n^{1/2}$ can be proved by mathematical induction (but it is way too late in the part of the world I am leaving to try to write it)


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