How to prоve the convergence of the sequence?

Question

I need to prove the convergence of $$x_n=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-2\sqrt{n+1}$$ I`m not very strong in this section of math. What should I do? Please help.

Answer 1

What about slicing the second term to form another sum, and regroup ?

By "detelescoping", we write

$$\sqrt{n+1}-1=\sum_{k=1}^n\left(\sqrt{k+1}-\sqrt{k}\right)$$

and study

$$\sum_{k=1}^n\left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right)=\sum_{k=1}^n\frac{2k+1-2\sqrt{k(k+1)}}{\sqrt k}\\ =\sum_{k=1}^n\frac1{\sqrt k\left(2k+1+2\sqrt{k(k+1)}\right)}.$$

The terms are of order $k^{-3/2}$, ensuring convergence.

Answer 2

We have \begin{align}x_n-x_{n-1}&=\frac1{\sqrt{n}}-2(\sqrt{n+1}-\sqrt{n}) \\&=\frac1{\sqrt{n}}-\frac2{\sqrt{n+1}+\sqrt{n}} \\&=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}(\sqrt{n+1}+\sqrt{n})} \\&=\frac1{\sqrt{n}(\sqrt{n+1}+\sqrt{n})^2} \\&\le\frac1{n^{3/2}} \end{align} so $\displaystyle\lim_{n\to\infty}=\sum^\infty_{n=1}(x_n-x_{n-1})$ is finite.

Answer 3

Let $(u_n)=(x_{n+1})-(x_n)$, for $n\in\Bbb N^*$, we have that \begin{align} u_n&={1\over\sqrt{n+1}}+2(\sqrt{n+1}-\sqrt{n+2})\\ &={1\over\sqrt{n+1}}-{2\over \sqrt{n+1}+\sqrt{n+2}}\\ &={1\over\sqrt{n+1}\ (\sqrt{n+1}+\sqrt{n+2})^2}\\ &=\mathcal{O}({1\over n\sqrt{n}}) \end{align} $(u_n)$ is a positive sequence, and $\sum{1\over n\sqrt{n}}$ is convergent, $\sum u_n$ is therefore convergent. Hence, $(x_n)$ is convergent.

Answer 4

As Jack D'Aurizio commented, it can be done in a million ways.

Since you already received good and simple answers, let me show another one.

$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=H_n^{\left(\frac{1}{2}\right)}$$ where appear generalized harmonic numbers. Using asymptotics, for large values of $n$, we have $$H_n^{\left(\frac{1}{2}\right)}=2 \sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac1 {2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ On the other hand $$\sqrt{n+1}=\sqrt{n}\sqrt{1+\frac{1}{n}}=\sqrt{n}\left(1+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)\right)=\sqrt{n}+\frac1 {2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ making $$x_n=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-2\sqrt{n+1}=\zeta \left(\frac{1}{2}\right)-\frac1 {2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ from which you can conclude.