What can be said about this sequence

[Natasha1]

What can be said about this sequence:

1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ... + \surd n

Hint: it might be helpful to consider \int_{a}^{b} \surd x dx for suitable a and b

 

[LeonhardEuler]

That's a pretty ambiguous question. Might they want you to show that it is bounded between two integrals with different limits? That's faily easy to do. But the question is so open that its hard to say that's what thery're looking for.

 

[Natasha1]

That's a pretty ambiguous question. Might they want you to show that it is bounded between two integrals with different limits? That's faily easy to do. But the question is so open that its hard to say that's what thery're looking for.

On this one my lecturer said it was an open question but what he wanted us to do is to say as much as we could about it.

 

[LeonhardEuler]

Well, for starters it is easy to see whether this series converges (n-th term test). As far as the integral goes, look at a proof of the integral test. It will probably use the fact that as long as the function is strictly increasing, then you can bound the series between two definite integrals. So you can bound this particular series between two integrals of the form \int_a^b\sqrt{x}dx for different a and b.

 

[StatusX]

If you want to see how fast the series grows, look at the integral of \sqrt{x} from 0 to 1. Approximate this by summing over strips of width 1/n to get an upper and lower bound, and this approximation will get better as n grows. You'll be able to find an expression containing n whose quotient with this sum has a limit of 1 as n goes to infinity. But, as was mentioned, the question is ambiguous, and I don't know if that's what you want.

 

[LeonhardEuler]

If you want to see how fast the series grows, look at the integral of \sqrt{x} from 0 to 1. Approximate this by summing over strips of width 1/n to get an upper and lower bound, and this approximation will get better as n grows. You'll be able to find an expression containing n whose quotient with this sum has a limit of 1 as n goes to infinity. But, as was mentioned, the question is ambiguous, and I don't know if that's what you want.

Are you sure [0,1] is the right interval to integrate over? It seems to me like you should be integrating from 0 to n or from 0 to n-1 to get an approximation of this series. But I may just not be seeing what you mean.

 

[Natasha1]

Are you sure [0,1] is the right interval to integrate over? It seems to me like you should be integrating from 0 to n or from 0 to n-1 to get an approximation of this series. But I may just not be seeing what you mean.

Guys is this the best way of takling the problem:

Let:

S_n= 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ... + \surd n

Then as \{\sqrt{r}, r=1,2, \dots,\ n\} is an increasing sequence:

\int_{x=1}^{n+1} \sqrt{x-1} dx

So doing the integrals gives:

\frac{2}{3}(n)^{3/2}<S_n<\frac{2}{3} ((n+1)^{3/2}-1)

Anything else I can say about this sequence? (just through anything relevant in, I would most appreciate it)

 

[Gokul43201]

Natasha : This is your third thread in this forum today, and judging from all of them, it appears that you may not have read the guidelines for posting in this forum.

Please read the sticky thread at the top of this sub-forum, or the second link in my signature.

We expect people to show their effort first, when asking for help. Clearly, from subsequent posts, you have shown this, but do so in the OP henceforth.

 

[Natasha1]

Does the series converge or diverge?

 

[LeonhardEuler]

Everything you did is right. See if you can use the nth term test to see if this series converges.

 

[StatusX]

Are you sure [0,1] is the right interval to integrate over? It seems to me like you should be integrating from 0 to n or from 0 to n-1 to get an approximation of this series. But I may just not be seeing what you mean.

You'll need to factor 1/\sqrt{n} out of the sum, which will leave the desired series. The only real difference is that this way makes it clearer that the expression actually converges to the same limit as the sum, since the region of integration is fixed.

 

[Natasha1]

Everything you did is right. See if you can use the nth term test to see if this series converges.

Leonhardeuler,

I used the nth term series test to prove that as the series tends to infinity as n--> infinity therefore the series diverges.

Is this ok?

 

[LeonhardEuler]

Yes, that works.