# Sequence proof (Hardish)

1. May 22, 2012

### sid9221

http://dl.dropbox.com/u/33103477/summands.png [Broken]

Even with the hint, I'm confused on what to use on this ? Any idea's ?

Last edited by a moderator: May 6, 2017
2. May 22, 2012

### sharks

The summand to compare with is most probably: 1/n

3. May 22, 2012

### gopher_p

Forget the hint for a second. It's a tad bit confusing (though you will see what they mean eventually).

Can you find an upper bound for each of the $x_n$? A lower bound?

If that's not enough to get you started ... can you establish a ... relation ... between each of the "terms" of the $x_n$?

4. May 22, 2012

### thrill3rnit3

$\frac{1}{n+1}$ = $\frac{1}{\sqrt{(n+1)^2}}$ = $\frac{1}{\sqrt{n^2+2n+1}}$ < $\frac{1}{\sqrt{n^2+2n}}$ < $\frac{1}{\sqrt{n^2}}$ = $\frac{1}{n}$

In short, we've sandwiched $\frac{1}{\sqrt{n^2+2n}}$ between $\frac{1}{n+1}$ and $\frac{1}{n}$

5. May 23, 2012

### gopher_p

OK. Not exactly what I was trying to get you to do, but we might be able to make it work.

Can you get similar bounds for the other terms? The $\frac{1}{\sqrt{n^2+2k}}$ for $1\leq k<n$?

While you're doing that, go ahead and figure out how many terms there are in $x_n$.

6. May 23, 2012

### deluks917

Doesn't the sequence diverge?

note: n^2 + 2n ≤ 4n^2

so 1/sqrt(n^2+ 2n) ≥ 1/sqrt(4n^2) = 1/(2n). But the sum of 1/(2n) diverges.

7. May 23, 2012

### sid9221

The question clearly states that it converges to 1.
So everything said till now makes no sense.

8. May 23, 2012

### deluks917

edit: see whats going on

Last edited: May 23, 2012
9. May 23, 2012

### Infinitum

Seconded. The series diverges. You can see it goes on increasing(>2) if you try with n>9, and the proof is right here.

Edit : I just realized my mistake. I took into consideration that the general term is

$T_n = \frac{1}{\sqrt{n^{2}+2n}}$

But the way the series is made, it will not have this as the general term. Gopher's method would work out best.

Last edited: May 23, 2012
10. May 23, 2012

### Karamata

I think that best solution (for me) is @gopher_p solution.

You know that:

$\frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+4}} + \cdots + \frac{1}{\sqrt{n^2+2n}}\le \frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+2}} + \cdots + \frac{1}{\sqrt{n^2+2}} = n\cdot \frac{1}{\sqrt{n^2+2}} =\frac{n}{\sqrt{n^2+2}}$

On the other hand

$\frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+4}} + \cdots + \frac{1}{\sqrt{n^2+2n}}\ge \frac{1}{\sqrt{n^2+2n}} + \frac{1}{\sqrt{n^2+2n}} + \cdots + \frac{1}{\sqrt{n^2+2n}} = n\cdot \frac{1}{\sqrt{n^2+2n}} =\frac{n}{\sqrt{n^2+2n}}$

So, you have

$\frac{n}{\sqrt{n^2+2n}} \le \frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+4}} + \cdots + \frac{1}{\sqrt{n^2+2n}}\le \frac{n}{\sqrt{n^2+2}}$

11. May 23, 2012

### sid9221

Guy's this is a past exam paper question and it has come up twice. If it was a mistake it wouldn't have come up two times in exactly the same form.

On a side note this has come up in the sequences portion of the paper(we have separate sections for different topics on the paper's easier bit), dunno if that's a hint.

12. May 23, 2012

### Infinitum

The question is correct, there is no mistake. Karamata almost gave you the solution, all you need to see is how the sandwich limits behave when n -> infinity.