# Finding $\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}{n^{3/2}}$

$$\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}{n^{3/2}}$$

How to find the limit? Any help will be appreciated.

By Stolz we have: $$\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}{n^{3/2}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}}{n^{3/2}-(n-1)^{\frac{3}{2}}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}\left(n^{3/2}+(n-1)^{\frac{3}{2}}\right)}{n^3-(n-1)^3}=$$ $$=\lim_{n\rightarrow\infty}\frac{n^2+\sqrt{n(n-1)^3}}{3n^2-3n+1}=\frac{2}{3}$$

• Looks like serial down voting. +1 to compensate – Paramanand Singh Feb 3 '17 at 5:14
• @Dr. MV In this forum sometimes we get very many up-votes for primitive questions and down-votes for beautiful solutions or problems. This is our life here. – Michael Rozenberg Feb 3 '17 at 5:16
• @MichaelRozenberg You're right. I've had such experiences. ;-) – Mark Viola Feb 3 '17 at 5:32

HINT:

$$\sum_{k=1}^n \frac{\sqrt k}{n^{3/2}}=\frac1n \sum_{k=1}^n \sqrt{\frac{k}{n}}$$

Now, think about a Riemann sum.

Alternatively, bound the sum by integrals as

$$\frac{1}{n^{3/2}}\int_1^{n+1}\sqrt{x}\,dx\le \sum_{k=1}^n \frac{\sqrt k}{n^{3/2}}\le \frac{1}{n^{3/2}}\sqrt{n}+\frac{1}{n^{3/2}}\int_1^{n}\sqrt{x}\,dx$$

and use the squeeze theorem.

• Why was this down voted?? Would the down voter care to comment? – Mark Viola Feb 3 '17 at 5:07
• The definite route to go for such limits is the Riemann sum. +1 – Paramanand Singh Feb 3 '17 at 5:14
• And downvoters please have some courage to write a comment. – Paramanand Singh Feb 3 '17 at 5:16
• @ParamanandSingh Yes, I agree. In the original post, the OP had mentioned the use of the squeeze theorem. So, I added the alternative. Any way ... thank you as always my friend! – Mark Viola Feb 3 '17 at 5:31