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

Question

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 06 '17 at 01:36
The title is about the sequence, and the quote box is about the series. — peterwhy, Jun 06 '17 at 01:37
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 06 '17 at 01:46
@Taladris I made a confusion of one theorem for sequences and used it for series. — Roland, Jun 06 '17 at 01:47
This search and this search in Approach0 return several similar questions. — Martin Sleziak, Jun 06 '17 at 04:02
Does this series diverge?, $\sum a_n = \sum{(\sqrt{n+1} - \sqrt{n})}$ Prove divergence, Does $\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}}$ converge?, Evaluate $\sum \sqrt{n+1} - \sqrt n$ — Martin Sleziak, Jun 06 '17 at 04:04

Zubin Mukerjee · Accepted Answer · 2017-06-06T01:54:21.013

13

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.

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

edited Jun 06 '17 at 01:54

answered Jun 06 '17 at 01:40

Zubin Mukerjee

17,832

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 06 '17 at 02:34
6

@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 06 '17 at 02:42

score 6 · Answer 2 · answered Jun 06 '17 at 01:46

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.

score 0 · Answer 3 · answered Jun 06 '17 at 02:36

0

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

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

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

answered Jun 06 '17 at 02:36

hamam_Abdallah

62,951

score -1 · Answer 4 · answered Dec 14 '22 at 21:43

-1

it is possible to show that $$\sum _{j=1}^{\infty } \left(\left(\sqrt{j+1}-\sqrt{j}\right)-\frac{\sqrt{\frac{1}{j}}}{2}\right)+2 \zeta \left(-\frac{1}{2}\right)+\frac{\zeta \left(\frac{1}{2}\right)}{2}=-\sqrt{2}$$ although mathematics numerically does not calculate it well, and therefore $$\sum _{j=1}^{\infty } \left(\sqrt{j+1}-\sqrt{j}\right)=-\sqrt{2}$$

answered Dec 14 '22 at 21:43

capea

456

I don't believe your definition of a convergent series is the same as the OP's. Can you elaborate on how you got that result? – Udi Fogiel Dec 14 '22 at 22:09
What you believe or stop believing is up to you, but mathematics is what it ia , voting negatively on an issue that is not understood is foolish but without wanting to be arrogant, it is something that you can look at WIKI about the analytical continuation. – capea Dec 15 '22 at 10:49

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

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