$$\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}{n^{3/2}}$$
How to find the limit? Any help will be appreciated.
Finding $\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}{n^{3/2}}$
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$\begingroup$
calculus
integration
limits
radicals
riemann-sum
2 Answers
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By Stolz we have: $$\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}{n^{3/2}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}}{n^{3/2}-(n-1)^{\frac{3}{2}}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}\left(n^{3/2}+(n-1)^{\frac{3}{2}}\right)}{n^3-(n-1)^3}=$$ $$=\lim_{n\rightarrow\infty}\frac{n^2+\sqrt{n(n-1)^3}}{3n^2-3n+1}=\frac{2}{3}$$
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0Looks like serial down voting. +1 to compensate – 2017-02-03
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0@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. – 2017-02-03
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0@MichaelRozenberg You're right. I've had such experiences. ;-) – 2017-02-03
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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.
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0Why was this down voted?? Would the down voter care to comment? – 2017-02-03
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0The definite route to go for such limits is the Riemann sum. +1 – 2017-02-03
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0And downvoters please have some courage to write a comment. – 2017-02-03
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0@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! – 2017-02-03