To be honest, I don't have much ideas on how to prove it. I only had an idea of using inequation $\frac{1}{n + 1} < \ln(1 + 1/n) < \frac{1}{n}$. But for sure it doesn't give us good approximation.
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1I assume there's supposed to be a square root in the last term there? – John Barber Nov 22 '20 at 01:49
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1Impossible. Left side is algebraic, right side transcendental. Do you mean a limit as $n\to\infty$? – Oscar Lanzi Nov 22 '20 at 01:51
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1This has been asked and answered before: https://math.stackexchange.com/q/2973156/42969. – Martin R Nov 22 '20 at 01:55
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@JohnBarber Thank you for noticing typo! – math-traveler Nov 22 '20 at 01:55
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4This is certainly a Riemann sum. – John Barber Nov 22 '20 at 01:57
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2This could help maybe. Put $$S_n=\dfrac{1}{\sqrt{n(n+1)}} + \dfrac{1}{\sqrt{(n + 1)(n+2)}} + \cdots + \dfrac{1}{\sqrt{2n(2n + 1)}}$$ One has $$\frac{1}{n+1}\lt \frac{1}{\sqrt{n(n+1)}}\lt\frac 1n$$ so $$\frac{1}{n+1}+\cdots+\frac{1}{2n+1}\lt S_n\lt\frac 1n+\cdots+\frac{1}{2n}$$ – Piquito Nov 22 '20 at 04:06