# Verify if $\sum(\sqrt{n+1}-\sqrt{n})$ is convergent or divergent

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?

• You proved that the sequence $a_n$ converges to zero. However, this does not imply that series $\sum a_n$ converges. In fact, the series diverges since it is comparable to $\sum \frac {1}{\sqrt n}$ which is divergent since it is a $p$-series with $p \le 1$. – User8128 Jun 6 '17 at 1:36
• The title is about the sequence, and the quote box is about the series. – peterwhy Jun 6 '17 at 1:37
• You should review your tests: having $b_n$ decreasing and converging to $0$ does not imply that $\sum b_n$ is convergent. The classic example is the harmonic series $\sum \frac{1}{n}$. The series $\sum \frac{1}{2\sqrt{n}}$ is another one. – Taladris Jun 6 '17 at 1:46
• This search and this search in Approach0 return several similar questions. – Martin Sleziak Jun 6 '17 at 4:02

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.

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

• 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. – Roland Jun 6 '17 at 2:34
• @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 :) – 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.