# Show that the sequence $x_n=\sum\limits_{k=1}^n\frac1{\sqrt{k+1}+\sqrt{k}}$ is unbounded. [duplicate]

Consider the sequence $$\{x_n\}_{n\ge1}$$ defined by $$x_n=\sum_{k=1}^n\frac{1}{\sqrt{k+1}+\sqrt{k}}, \forall n\in\mathbb{N}.$$ Is $$\{x_n\}_{n\ge 1}$$ bounded or unbounded.

I solved the problem as stated in the answer posted by me. Is it possible to solve the problem in a more better and rigorous manner?

Here goes the solution I found out. Let $$a_k$$ be defined as $$a_k=\frac{1}{\sqrt{k}+\sqrt{k+1}}, \forall k\in\mathbb{N}.$$ Then, $$x_n=\sum_{k=1}^n a_k, \forall n\in\mathbb{N}.$$ Now $$a_k=\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}=\sqrt{k+1}-\sqrt{k}, \forall k\in\mathbb{N}.$$

This implies that $$x_n=\sqrt{n+1}-1, \forall n\in\mathbb{N}.$$ Now intuition says that $$\{x_n\}_{n\ge 1}$$ is an unbounded sequence. But let us try to prove it rigorously.

Let us assume that $$\{x_n\}_{n\ge 1}$$ is bounded above. This implies that, we can find $$M\in\mathbb{R}$$ such that $$x_n\le M, \forall n\in\mathbb{N}$$. Now let $$\lceil M\rceil=n_1\implies M\le n_1.$$

Now $$x_{n_1^2+4n_1+3}=\sqrt{n_1^2+4n_1+3+1}-1=\sqrt{(n_1+2)^2}-1=n_1+1.$$

This implies that $$x_{n_1^2+4n_1+3}=n_1+1>n_1\ge M\implies x_{n_1^2+4n_1+3}>M.$$ But we have assumed that $$x_n\le M, \forall n\in\mathbb{N}.$$ Contradiction.

This implies that $$\{x_n\}_{n\ge 1}$$ is not bounded above, which in turn implies that $$\{x_n\}_{n\ge 1}$$ is not bounded.

$$\sqrt {k+1}+\sqrt k \leq \sqrt {2k}+\sqrt k<3\sqrt k$$. Hence the given sum is at least $$\sum\limits_{k=1}^{n} \frac 1 {3\sqrt k}$$ Now use the fact that the series $$\sum\limits_{k=1}^{n} \frac 1 {3\sqrt k}$$ is divergent.

• I have a doubt. Is it true that, every divergent series is unbounded? From what you have stated we can surely say that the series $\{x_n\}_{n \ge 1}$ is divergent, but can we state that it is unbounded. Like, I have a counter example. Consider the series $1-1+1-1+...-1+1-...$. We know that this series is divergent. But the partial sums of this series are either $-1$ or $+1$, which implies that the partial sums of this series are bounded ($-1$ being the lower bound and $+1$ being the upper bound). – Eduline Jan 26 at 11:09
• @SanketBiswas A series of positive terms is divergent iff the partial sums tend to $+\infty$ in which case the partial sums form an unbounded sequence. – Kavi Rama Murthy Jan 26 at 11:25
• Okay thanks, I get it now! – Eduline Jan 26 at 11:26