# Finding limit of a series.

I need to find $$\lim_{n\rightarrow \infty} \frac {\sqrt{1}+\sqrt{2}+...+\sqrt{n}} {n^{\frac 4 3}}$$ I know from numerical calculation that this approaches infinity, but I dont know how to prove it (I need to do so using only infinite series, not integrals or anything of the sort). I tried comparing it to a smaller expressions but all such expressions approach $$0$$ or a constant when I tried, so I'm stuck and would appreciate help.

• Compare the numerator with $\int_0^n x^{1/2}\,dx$. – Angina Seng Dec 9 '19 at 18:27

Looking at $$\sum_{i=1}^n\sqrt{n}$$, and taking the upper $$\text{floor}(n/2)$$ terms, they are all greater than or equal to $$\sqrt{(n-1)/2}.$$ This gives us

$$\frac{\sqrt{1}+\ldots+\sqrt{n}}{n^{4/3}}>\frac{[\sqrt{(n-1)/2}]\text{floor}(n/2)}{n^{4/3}}\geq\frac{[\sqrt{(n-1)/2}][(n-1)/2]}{n^{4/3}}=\frac{1}{2^{3/2}}\frac{(n-1)^{3/2}}{n^{4/3}}.$$ The Right most term converges to infinity, hence so does the left most.

Edit: Replaced $$(n-1)/2$$ with $$\text{floor}(n/2)$$, because $$n$$ can be even. In case you're not familiar, $$\text{floor}(n/2)$$ is just the greatest integer less or equal to $$n/2.$$ So it's either $$n/2$$ or $$(n-1)/2$$.

$$\lim_{n\to \infty} \frac{\sqrt{n}}{(n)^{\frac{3}{2}} - (n-1)^{\frac{3}{2}}} = \frac{2}{3}$$

Thus,

$$\lim_{n\to \infty}\frac{\sqrt{1}+\sqrt{2}+...+\sqrt{n}}{n^{\frac{3}{2}}} = \frac{2}{3}$$

And I suppose you can finish the rest of the proof.

By Stolz-Cesaro we have

$$\frac{a_n}{b_n}=\frac {\sqrt{1}+\sqrt{2}+...+\sqrt{n}} {n^{\frac 4 3}} \implies \frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\sqrt{n+1}}{(n+1)^{\frac 4 3}-n^{\frac 4 3}}\to \infty$$

indeed

$$(n+1)^{\frac 4 3}-n^{\frac 4 3}=\frac{(n+1)^4-n^4}{(n+1)^\frac83+(n(n+1))^\frac43+n^\frac83}\sim\frac{4n^3}{3n^\frac83}=\frac43n^\frac13$$

We can rewrite the limit as a sum

$$\lim_{n\to\infty} \frac{\sum_{k=1}^n \sqrt{k}}{n^{\frac{4}{3}}} = \lim_{n\to\infty} n^{\frac{1}{6}} \sum_{k=1}^n \sqrt{\frac{k}{n}}\cdot\frac{1}{n}$$

The term on the right is a Riemann sum that converges to

$$\int_0^1 \sqrt{x}dx = \frac{2}{3}$$

Thus the limit asymptotically goes as $$\sim \frac{2}{3} n^{\frac{1}{6}}$$, which makes it infinite as $$n\to\infty$$