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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?

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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.

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$\sqrt {k+1}+\sqrt k \leq \sqrt {2k}+\sqrt k<3\sqrt k$. 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.

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  • $\begingroup$ 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). $\endgroup$ – Eduline Jan 26 at 11:09
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Jan 26 at 11:25
  • $\begingroup$ Okay thanks, I get it now! $\endgroup$ – Eduline Jan 26 at 11:26

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