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

[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.

I do hereby request you gentlemen to offer your advices.
∑n=1∞[n+1−n]

 

[Office_Shredder]

It's not cauchy. If you pick any n, you can always pick k large enough that that difference is bigger than 1.

 

[Hall]

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?

 

[pasmith]

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

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

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

 

[Hall]

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.

 

[pasmith]

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 [itex]\sqrt{n + k + 1} - \sqrt{n + 1} > 1[/itex]; shift [itex]\sqrt{n+1}[/itex] to the right hand side and then square both sides.

 

[WWGD]

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

 

[Office_Shredder]

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##.

 

[WWGD]

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 ;) ).

 

[SammyS]

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

 

[Hall]

Start from [itex]\sqrt{n + k + 1} - \sqrt{n + 1} > 1[/itex]; shift [itex]\sqrt{n+1}[/itex] 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.

 

[Hall]

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.

 

[Hall]

@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 .

 

[Office_Shredder]

##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.

 

[epenguin]

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).

 

[WWGD]

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.

 

[SammyS]

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.

 

[epenguin]

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