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# Can anyone......

Find the sum of the finite series

$\frac{\sqrt{1} +\sqrt{2} +\sqrt{3} + \sqrt{4}+............ \sqrt{n}}{n^{2}}$,

with $$n$$ being any positive integer?

Note by Bryan Lee Shi Yang
2 years, 3 months ago

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$\frac23(n+1)^{3/2}-\frac12(n+1)^{1/2}-\frac12\le\sqrt1+\dotsb+\sqrt n\le\frac23(n+1)^{3/2}-\frac12(n+1)^{1/2}$

In other words, $$\sqrt1+\dotsb+\sqrt n=\frac23(n+1)^{3/2}-\frac12(n+1)^{1/2}-\text{error}$$, where "error" is between 0 and 1/2.

Source · 2 years, 2 months ago