Verify if the series $\sum a_n$ with $a_n=\sqrt{n+1}-\sqrt{n}$ is convergent or divergent.

What I did is

$$a_n=\sqrt{n+1}-\sqrt{n}\times\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt {n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$ $$<\frac{1}{2\sqrt{n}}=b_n$$

Since $b_n$ is monotone decreasing and $b_n\rightarrow 0$ when $n\rightarrow \infty$ then $b_n$ is convergent.

Using the comparison test, we have that $0\leq a_n\leq b_n$. If $b_n$ is convergent then $a_n$ is convergent.

Is it right?


The series diverges. This can be seen using the fact that it is a "telescoping series".

The $k^\text{th}$ partial sum can be written as follows:

\begin{align} S_k &= \displaystyle\sum_{n=1}^k \Big(\sqrt{n+1} - \sqrt{n}\Big)\\\\ &= \big(\sqrt{2} - \sqrt{1}\,\big) + \big(\sqrt{3} - \sqrt{2}\,\big) + \big(\sqrt{4} - \sqrt{3}\,\big) + \cdots +\big(\sqrt{k+1} - \sqrt{k}\,\big)\\\\ &=-\sqrt{1} + \big(\sqrt{2} -\sqrt{2}\,\big) + \big(\sqrt{3} -\sqrt{3}\,\big) + \big(\sqrt{4} - \cdots -\sqrt{k}\,\big) + \sqrt{k+1}\\\\ &=\sqrt{k+1} - 1\\ \end{align}

As $k$ goes to infinity, this diverges, so the infinite sum does not converge.

$$\displaystyle\sum_{n=1}^\infty \Big(\sqrt{n+1} - \sqrt{n}\Big) = \lim_{k\to\infty} S_k = \infty$$

  • $\begingroup$ Interesting, so if the kth partial sum diverges then the infinite series diverges? The theorem that I found just say about the convergence, then I'm not sure if it's valid for divergence. $\endgroup$ – Roland Jun 6 '17 at 2:34
  • 3
    $\begingroup$ @Roland An infinite sum is defined as the limit of the partial sums. So when I say the $k^\text{th}$ partial sum diverges as $k$ goes to infinity, that is actually by definition what it means for the infinite sum to diverge :) $\endgroup$ – Zubin Mukerjee Jun 6 '17 at 2:42

In the same manner,

$$a_n>\frac1{\sqrt{n+1}+\sqrt{n+1}}=\frac12\frac1{\sqrt {n+1}}>\frac12\frac1{\sqrt{n+n}}=\frac1{2\sqrt2}\frac1{\sqrt n}=b_n$$

By the comparison test, we easily see that

$$\sum_{n=1}^\infty a_n>\sum_{n=1}^\infty b_n\to+\infty$$

which follows from the p-series.


$$\sqrt {n+1}-\sqrt {n}=$$

$$\sqrt {n}(\sqrt {1+\frac {1}{n}}-1)$$

$$\sim \sqrt {n}.\frac {1}{2n} $$ $$\sim \frac {1}{2\sqrt {n}} $$ the series diverges.


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