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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, 1 month 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 Akiva Weinberger · 2 years ago

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