$x_n=1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt n}-2\sqrt n$

Question

$x_n=1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt n}-2\sqrt n$

From here Investigating the convergence of a series using the comparison limit test, Part II I can see that the sequence converges. I was trying to use the monotone convergence theorem instead since $x_{n+1}-x_n=\frac{-1}{\sqrt {n+1}(\sqrt n+\sqrt{n+1})^2}\leq0$. How should I go about guessing the lower bound and using induction and MCT to show convergence?

Answer 1

$$2(\sqrt n - \sqrt{n-1})=\frac2{\sqrt n+\sqrt{n-1}}\geq\frac1{\sqrt n}\geq\frac2{\sqrt n+\sqrt{n+1}}=2(\sqrt{n+1}-\sqrt n)$$ So we have $$2\sqrt n\geq\sum_{k=1}^n\frac1{\sqrt k}\geq2(\sqrt{n+1}-1)$$ You only need the lower bound, but it's nice to have an upper bound as well.

Answer 2

Since $1/\sqrt{x}$ is a decreasing function, $$ 1+\frac{1}{\sqrt2}+\dots+\frac{1}{\sqrt n}\ge\int_{1}^{n+1}\frac{dx}{\sqrt x}=2\sqrt{n+1}-2. $$