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 (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.

 

[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 \sqrt{n + k + 1} - \sqrt{n + 1} > 1; shift \sqrt{n+1} 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 \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.

 

[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.