Use integration to find the sum of a series

raopeng
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Homework Statement


Find the sum using integration: lim_{n→∞} \frac{n}{(n+1)^2} + ... + \frac{n}{(2n)^2}


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The Attempt at a Solution


I think this requires a clever construction of a series of an finite integral which after integration gives the series. Then it can be solved by summing up the series inside the integral then integrate the whole thing. But now what is bothering me is how to construct such a function, every function I have tried seems remotely far from the series in the question. Thanks for your time.
 
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How about bounding the sum by a pair of integrals over (n, 2n)?
 
Or show that the given sum is equivalent to a Riemann sum for some function. So taking the limit is equivalent to finding the integral.
 
Oh thanks so much!
 
I transform the series in this way: Ʃ_{k=1} ^{n} \frac{1}{(1+k/n)^2 n} which turns into an integral \int^1 _0 \frac{dx}{(1+x)^2}. An integration gives 1/2.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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