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find this limit: $$\displaystyle\lim_{n\to+\infty}\left[\sum_{k=1}^{n}\left(\dfrac{1}{\sqrt{k}}- \int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt\right)-2\sqrt{n}\right]$$

Git Gud
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math110
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    @math110 You've asked 21 questions on here and accepted exactly $0$ answers. Please consider going through your questions and accepting your favourite answers to questions in which you got at least one satistying answer. – Git Gud Mar 20 '13 at 07:02
  • sorry,integrate from 0 to $\dfrac{1}{\sqrt{i}}$ – math110 Mar 20 '13 at 07:04
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    You have asked 17 questions during the last week and provided NOT A SINGLE INDICATION about what you know, what you tried, etc. Must be some kind of experiment to see how far one can push the system. – Did Mar 20 '13 at 09:17

2 Answers2

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First, you can check that $$ \int_0^{1/\sqrt k} \frac{t^2}{1+t^2}\,dt=\int_0^{1/\sqrt k} 1-\frac{1}{1+t^2}\,dt=\frac{1}{\sqrt k}-\arctan\frac{1}{\sqrt k}, $$ so you may rewrite your limit as $$ \lim_{n\to\infty}\left[\sum_{k=1}^n\left(\arctan\frac{1}{\sqrt k}\right)-2\sqrt n\right]. $$

This is at least somewhere to start.

Additionally, we have $\arctan x + \arctan y = \arctan\frac{x + y}{1 - xy}$. This provides some hope of being able to simplify the sum on the inside.

Further edit: No guarantee this will pan out to work, but it does seem more tractable.

Ian Coley
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We have $$\int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt=\frac{1}{\sqrt k}-\arctan\frac{1}{\sqrt k}$$ Now if we denote by $$u_n=\sum_{k=1}^n\arctan\frac{1}{\sqrt k}-2\sqrt{n}$$ we have $$u_n-u_{n-1}=\arctan\frac{1}{\sqrt n}-2\sqrt{n}+2\sqrt{n-1}\sim\frac{-7}{12n\sqrt{n}},$$ and since the series $\sum\frac{1}{n\sqrt{n}}$ is convergent then the sequence $(u_n)$ is also convergent to say $\ell$ and we we have $$\sum_{k=n+1}^\infty u_k-u_{k-1}=\ell-u_n\sim\frac{-7}{12}\sum_{k=n+1}^\infty\frac{1}{k\sqrt{k}}\sim\frac{-7}{12}\int_{n+1}^\infty\frac{dx}{x\sqrt{x}}=\frac{-7}{6\sqrt{n}}$$ so we find the asymptotic equality $$u_n=\ell+\frac{7}{6\sqrt{n}}+o(\frac{1}{\sqrt{n}})$$