Find $\lim\limits_{n \to \infty} \frac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$.

$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$$

$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}} =\lim_{n\to \infty} \dfrac1{n}\sum^{n}_{k = 1} \sqrt{\dfrac k n}$$

While searching this question I found, Turning infinite sum into integral.

Like in the accepted answer I first compared my series to LRAM,

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

I got $a = 0$, $b = 1$ and $f(x) =\sqrt{x}$ so,

$$\int_0^1 \sqrt{x}\ dx = \dfrac2 3$$ should be the answer.

Is there any simpler method to do this sum ? I have not learnt this method to do infinite sums so I can't use it.

• You haven't learned this method to do infinite sums, but have you learned about integrals and their definition? I have seen questions like this appear on the various standardized tests, and the intent is to recognize them as integrals and use an antiderivative to most quickly determine their value. Sep 3, 2017 at 21:12
• Jan 15, 2020 at 6:02

By the Stolz-Cesaro Theorem, one has \begin{eqnarray} &&\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}\\ &=&\lim_{n \to \infty} \dfrac{\sqrt{n+1}}{(n+1)\sqrt{n+1}-n\sqrt n}\\ &=&\lim_{n \to \infty} \dfrac{\sqrt{n+1}}{(n+1)\sqrt{n+1}-n\sqrt n} \dfrac{(n+1)\sqrt{n+1}+n\sqrt n}{(n+1)\sqrt{n+1}+n\sqrt n}\\ &=&\lim_{n \to \infty} \dfrac{\sqrt{n+1}[(n+1)\sqrt{n+1}+n\sqrt n]}{(n+1)^3-n^3}\\ &=&\frac23. \end{eqnarray}

• +1, I hate it when people beat me due to their fast typing speed. :P Sep 3, 2017 at 19:48
• If yours is independent, submit it anyway. Your explanation may be clearer. Sep 3, 2017 at 20:18
• @martycohen, If xpaul took stolz cesaro from me then what.... I will still submit my answer. :P Sep 3, 2017 at 20:31
• He did not take it away. This is not a zero sum game. Sep 3, 2017 at 20:42

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$$\frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; (n+1) \sqrt {n+1} \; - \; 1 \right) \; < \; \frac{2}{3} \left( \; (n+1) ( 1 +\sqrt n) \; - \; 1 \right) = \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n \right)$$ $$\frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n \right)$$

For anyone worried about the little estimate above, $$n + 1 < n + 2 \sqrt n + 1,$$ $$\sqrt {n+1} \; \; < \; \; 1 + \sqrt n.$$

• Thank you for the answer. Sep 4, 2017 at 11:06

Learning new techniques is good:

Method to express the infinite series as definite integral:

$1.$ Express the given series in the form $\sum\frac{1}{n}f(\frac{r}{n})$.

$2.$ Then the limit is its sum when $n\to \infty$, i.e, $\lim_{n\to\infty}\sum\frac{1}{n}f(\frac{r}{n})$

$3.$ Replace $\frac{r}{n}$ by $x$ and $\frac{1}{n}$ by $dx$ and $\lim_{n\to\infty}\sum$ by $\int$

$4.$ The upper and lower limit are limiting values of $\frac{r}{n}$ for first and last term of $r$ respectively.

For instance: $\sum_{r=1}^n=\int\frac{1}{n}f(\frac{r}{n})=\int_0^1f(x).dx$.

$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}=\lim_{n \to \infty} \dfrac{1}{n}\sum_{r=1}^n\sqrt{\frac{r}{n}}=\int_0^1\sqrt{x}=\frac{2}{3}$.