Rearranging the alternating harmonic series to sum to √2

[andrey21]

so that would give: n SQRT(n + 3) + 1

correct?

 

[micromass]

No, that would correspond to

$$n(\sqrt{n+3}+1)=\sqrt{n^3+3n^2}+n$$

 

[andrey21]

Hmm i a little confused as to how u got the above? I mite need little more help factoring n.

 

[micromass]

You do know that $$\sqrt{n^2a}=n\sqrt{a}$$ right? That's all I did...

 

[andrey21]

Ah i actually didnt know that! so with that in min I obtain:

n SQRT(1+3/n) + 1

Am I almost there??

 

[micromass]

Ah, yes, that is correct. So our limit is

$$\lim_{n\rightarrow +\infty}{\frac{3n}{n(\sqrt{1+3/n}+1)}$$

Cancel the n. Then substitute $$n=+\infty$$, and see what you get...

 

[andrey21]

Yes i see so would be:

3/SQRT(1+3/n) +1

So as n tends to infinty the 3/n becomes zero leaving:

3/SQRT(1) + 1

which is 3/2 or 1.5
Correct?

 

[micromass]

Yes, you'll get 3 in the numerator. But what happens if you substitute infinity in the denumerator?

 

[andrey21]

Is post 67 correct micromass?

 

[micromass]

Yes, 3/2 is the correct answer.

 

[andrey21]

fantastic I just have to use the binomial therom now it says to prove this. what part would i apply this to?

 

[andrey21]

Think I have solved it:

Using (1+x)^r where r is not a whole number. I obtain:

1+3/2.(n) - 1/8 - 1/16 ...

so as the expansion continues the values get closer to zero leaving 1 + 3/2(n)

 

[andrey21]

I realize I may have ae a mistake. Am i on the rite track?

 

[micromass]

fantastic I just have to use the binomial therom now it says to prove this. what part would i apply this to?

Use the binomial theorem to prove what??

 

[andrey21]

the limit I think.

 

[micromass]

No, I'm just confused. What exactly are you trying to prove now?

 

[andrey21]

My original question stated. Using simple algebra find the limit of this sequence as n tends to infinity. Then confirm this using the binomial therom.

 

[micromass]

I have no idea what they mean with "confirm with the binomial theorem"...

Do they mean this:

$$(x-y)^r=\sum_{k=0}^{+\infty}{\binom{r}{k}x^{r-k}y^k}$$

if so, you just need to substitute x=n², y=3n and r=1/2...

 

[andrey21]

Possibly, if that is the case I just sub in values for x y and r. Then what?

 

[micromass]

Calculate $$(n^2-n)^{1/2}-n$$ with the binomial theorem. What does that give you?

 

[andrey21]

Sorry I am nt to sure how to do that when r = 1/2

 

[andrey21]

Ok well I am going to come back to that. The bext question I am asked is to describe the sequence that SQRT(n^2 +3n) -n generates:

Substituting in values for n I obtain:

0,1,((SQRT10)-2),((SQRT18)-3), ((SQRT28) -4),...

From this am i correct in saying it is positive and monotonic, And converges to 3/2 as already discovered. Are there any upper lower bounds?

 

[micromass]

You'll have to prove that it is positive and monotonic...

 

[andrey21]

Ok well monotonic is where the next value is greater than or equal to the previous one correct?? So can't I just set two values next to each other:

1<((SQRT10) -2)

 

[micromass]

Yes, but this only proves it for n=1 and n=2. You'll need to show it for every n. You'll need to show that for every n

$$\sqrt{n^2+3n}-n\leq \sqrt{(n+1)^2+3(n+1)}-(n+1)$$

 

[andrey21]

Ah ok so from what u have written above. As the sequence is increasing we can say:
an + (an+1) is always positive.

so an + (an+1) = SQRT(n^2 +3n)-n + SQRT((n+1)^2 +3(n+1)) -(n+1)

 

[micromass]

No, you need to prove that...

 

[andrey21]

Scratch that above should it be an+1 - an is always positive??
so I will have:

an+1 - an = + SQRT((n+1)^2 +3(n+1)) -(n+1) - SQRT(n^2 +3n)-n

 

[micromass]

Yes.

(note: the forum rules explicitely forbid that you edit posts that already have been answered to. So please do not do this)

 

[andrey21]

Rite so is that just the answer then??