How Can I Find The Value Of This Limit 10 find this limit:
$$\displaystyle\lim_{n\to+\infty}\left[\sum_{k=1}^{n}\left(\dfrac{1}{\sqrt{k}}- \int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt\right)-2\sqrt{n}\right]$$
 A: 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.
A: 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
$$u_n=\ell+\frac{7}{6\sqrt{n}}+o(\frac{1}{\sqrt{n}})$$
