# Doubt regarding the series ##\sum [\sqrt{n+1} - \sqrt{n}\,]##

• Hall
In summary: Yes, the difference between a number and it’s square root can be made larger and...That would work, but it would require more work.
Hall
Homework Statement
We have to discuss the convergence of ##\sum_{n=1}^{\infty} [\sqrt{n+1} - \sqrt{n}\,]##.
Relevant Equations
Cauchy's Criterion says, for the sequence of partial sums ##(s_n)##, if for every ##\varepsilon \lt 0## there exits ##N##, such that for ##m, n \gt N##
$$| s_n - s_m| \lt \varepsilon$$
then the series converges.
We're given the series ##\sum_{n=1}^{\infty} [ \sqrt{n+1} - \sqrt{n} ]##.
##s_n = \sqrt{n+1} - 1##
##s_n## is, of course, an increasing sequence, and unbounded, given any ##M \gt 0##, we have ##N = M^2 +2M## such that ##n \gt N \implies s_n \gt M##. Thus, the series must be divergent.

But ##(s_n)## is a Cauchy sequence too, for ##m= n+k##, we have
## |s_{n+k} - s_n| = \sqrt{ n+k+1} - \sqrt{n+1}##
## | s_{n+k} - s_n| = \frac{k}{ \sqrt{n+k+1} + \sqrt{n+1} }##
By increasing ##n## we can make the above difference as small as we please. Thus, the series must convergent.

∑n=1∞[n+1−n]

Last edited by a moderator:
It's not cauchy. If you pick any n, you can always pick k large enough that that difference is bigger than 1.

Hall, Orodruin and topsquark
Office_Shredder said:
It's not cauchy. If you pick any n, you can always pick k large enough that that difference is bigger than 1.
I was trying to show we can make the difference small enough by taking large n, as n occurs in the denominator.

Can you please give a start for proving the divergence for this sequence using Cauchy’s criterion?

Hall said:
But ##(s_n)## is a Cauchy sequence too, for ##m= n+k##, we have
## |s_{n+k} - s_n| = \sqrt{ n+k+1} - \sqrt{n+1}##
## | s_{n+k} - s_n| = \frac{k}{ \sqrt{n+k+1} + \sqrt{n+1} }##
By increasing ##n## we can make the above difference as small as we please. Thus, the series must convergent.

To prove that $(s_n)$ is Cauchy, you must show that for all $\epsilon > 0$, there exists $N$ such that for all $m > N$ and all $n > N$ we have $|s_m - s_n| < \epsilon$.

Without loss of generality, you may assume $m > n$ and set $m = n + k$. But then if you increase $n$ for fixed $k$, you are also increasing $m$ by the same amount. This doesn't show that $|s_m - s_n|$ is bounded for all sufficiently large $m$ and $n$.

If instead you fix $n$ and increase $k$, you can make $|s_{n+k} - s_n| > 1$ as @Office_Shredder notes.

Well, I guess the sequence
$$a_k= s_{n+k} - s_n = \sqrt{n+k+1} -\sqrt{n+1}$$
Is an increasing sequence in ##k##. So, the difference will not decrease or remain below any given number, hence it is not Cauchy.

I don’t know, maybe due to cloudiness, why I’m unable to pick a k in order to make the difference greater than 1.

Hall said:
I don’t know, maybe due to cloudiness, why I’m unable to pick a k in order to make the difference greater than 1.

Start from $\sqrt{n + k + 1} - \sqrt{n + 1} > 1$; shift $\sqrt{n+1}$ to the right hand side and then square both sides.

Hall
IIRC, a telescope in ##\{ a_n \}##converges iff ## a_n ## itself converges.

Hall
Hall said:
Well, I guess the sequence
$$a_k= s_{n+k} - s_n = \sqrt{n+k+1} -\sqrt{n+1}$$
Is an increasing sequence in ##k##. So, the difference will not decrease or remain below any given number, hence it is not Cauchy.

I don’t know, maybe due to cloudiness, why I’m unable to pick a k in order to make the difference greater than 1.
We don't need to be careful about it. k=n clearly does the trick if ##n>10##.

Hall
Hall said:
Well, I guess the sequence
$$a_k= s_{n+k} - s_n = \sqrt{n+k+1} -\sqrt{n+1}$$
Is an increasing sequence in ##k##. So, the difference will not decrease or remain below any given number, hence it is not Cauchy.

I don’t know, maybe due to cloudiness, why I’m unable to pick a k in order to make the difference greater than 1.
How about ##k =n^2+n ##. This will work for even small n. Then show the difference is an increasing function., so that it holds for all larger n. Or you can complete the expression for a higher power ( And I'm not talking religion here either ;) ).

Hall
@Hall ,
Have you considered the possibility that the Series does not converge ?

pasmith said:
Start from $\sqrt{n + k + 1} - \sqrt{n + 1} > 1$; shift $\sqrt{n+1}$ to the right hand side and then square both sides.
##n+k+1 \gt 1 + n+1 - 2 \sqrt{n+1}##
##k \gt 1-2 \sqrt{n+1}##
Actually, I thought as 1 was recommended to me for a lower bound the value for k would have been like butter.

WWGD said:
How about ##k =n^2+n ##. This will work for even small n. Then show the difference is an increasing function., so that it holds for all larger n. Or you can complete the expression for a higher power ( And I'm not talking religion here either ;) ).
Yes, the difference between a number and it’s square root can be made larger and larger.

WWGD
SammyS said:
@Hall ,
Have you considered the possibility that the Series does not converge ?
I proved the divergence of series using monotone sequence method in the original post .

Hall said:
##n+k+1 \gt 1 + n+1 - 2 \sqrt{n+1}##
##k \gt 1-2 \sqrt{n+1}##
Actually, I thought as 1 was recommended to me for a lower bound the value for k would have been like butter.
There shouldn't be a minus sign here.

The original suggestion of 1 was picked out of thin air, there's no reason it will give a nice looking choice for k.

Am I confused or else can't this sum be written as a first plus a last term plus a lot of zeros (which would be divergent).

epenguin said:
Am I confused or else can't this sum be written as a first plus a last term plus a lot of zeros (which would be divergent).
Yes, some call it a telescoping sum, or telescope, though not sure why, but it converges if the nth term itself does.

WWGD said:
Yes, some call it a telescoping sum, or telescope, though not sure why, but it converges if the nth term itself does.
As to why it's called "telescoping":

See the following Wikipedia Link: https://en.wikipedia.org/wiki/Telescoping_(mechanics) which also has a Link to telescoping cylinder.

WWGD
I apologise, I did miss the point of this thread- my point was already in #1 of the OP.

## 1. What is the series ##\sum [\sqrt{n+1} - \sqrt{n}\,]## and how is it calculated?

The series ##\sum [\sqrt{n+1} - \sqrt{n}\,]## is a mathematical series that involves taking the square root of consecutive numbers and subtracting them from each other. It can be calculated using various methods, such as the sum of a geometric series formula or by using a calculator.

## 2. What is the purpose of this series and how is it used in science?

This series is often used in mathematics and physics to understand the behavior of sequences and series. It can also be used to approximate the value of certain integrals and to study the convergence of other series.

## 3. Is there a specific value or limit for this series?

Yes, the series ##\sum [\sqrt{n+1} - \sqrt{n}\,]## has a limit of 0 as n approaches infinity. This means that as n gets larger and larger, the terms in the series get closer and closer to 0, making the overall sum approach 0 as well.

## 4. How is this series related to the concept of convergence?

This series is an example of a convergent series, meaning that it has a finite sum. The concept of convergence in mathematics refers to the behavior of a sequence or series as its terms approach a specific value or limit.

## 5. Are there any real-world applications of this series?

Yes, this series has real-world applications in various fields such as finance, engineering, and economics. It can be used to model the growth of populations, the decay of radioactive materials, and the value of investments over time.

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