Solving Very Difficult Limit Homework

[dirk_mec1]

Homework Statement

http://www.sosmath.com/CBB/latexrender/pictures/f45d5e5b84cb5aecef3fd2f568739e32.gif

Homework Equations

hint:

http://www.sosmath.com/CBB/latexrender/pictures/fee1863dab606636f986e63c261b565f.gif
http://www.sosmath.com/CBB/latexrender/pictures/61d6e1cc712fcb6ac6f19b40851e85be.gif
http://www.sosmath.com/CBB/latexrender/pictures/ae169a7ff6b1d2c09a977e5aad223650.gif
http://www.sosmath.com/CBB/latexrender/pictures/cd14347d968bdb5be67f72e832558702.gif

The Attempt at a Solution

I don't understand how the hint helps in this limit. Could someone point me in the right direction?

 

[Billy Bob]

I didn't finish it, but here's my possible hint. If L=your limit, consider log L.

Edited to add: f(x) will be log(1+x). I "feel" like this is right.

 

[dirk_mec1]

Yes, that's helps a little bit but what do you mean by f(x) will be 1+x? How is that helpful?

 

[Billy Bob]

what do you mean by f(x) will be 1+x?

It will turn out that you will want to use f(x)=log(1+x).

Anyway, I did finish it now, and it works out. Use the log of the limit idea as I first suggested and apply properties of logs. At some step you will get terms with log(n+k). Rewrite as log[n(1+k/n)]=log n + log(1+k/n). Eventually you'll have to decide what to use for x, and it's not too hard to guess. Write out what you can, and I bet you will get it. If not, post your work and I'll give you more hints. I predict from time to time you'll say things to yourself like "wow, I can't believe those terms canceled out exactly, this must be right."

 

[dirk_mec1]

$$\ln(L) = \sum_{k=0}^n \frac{1}{2^n} \left( \stackrel{n}{k} \right) \left( \ln(n) + \ln \left(1+\frac{k}{n} \right) \right) - \ln(n)$$
Now if $$f(x) = \ln(1+x)$$ then B_n(1) is what I have, I only need to add an extra (1/2^n)* ln(2) and (subtract it later) to use the hint but for B_n(1)$$\ \sum_{k=0}^n \left( \stackrel{n}{k} \right) \ln \left(1+\frac{k}{n} \right) - \ln(2) \leq \frac{1}{2 \sqrt{n}} \leq \frac{1}{2}$$
$$\left( \frac{1}{2} \right) ^n \sum_{k=0}^n \left( \stackrel{n}{k} \right) \ln(n) = \ln(n)$$

so this cancels the last term leaving only that annoying thing from above plus that term which I needed to substract:
$$\left( \frac{1}{2} \right) ^n \left[ \frac{1}{2} + \ln(2) \right]$$So the limit is 1? Am I going in the right direction?

 

[Billy Bob]

Looking good, but B_n(1) would be 0. I used a different value for x.

(Also, leave $$\frac{1}{2\sqrt{n}}$$ as is. Don't replace it by 1/2.)

 

[dirk_mec1]

x=0? then

$$\ \sum_{k=0}^n \left( \stackrel{n}{k} \right) \ln \left(1+\frac{k}{n} \right) - \ln(1) \leq \frac{1}{2 \sqrt{n}}$$

This results in:

$$\left( \frac{1}{2} \right) ^n \left[ \frac{1}{2\sqrt{n}}\right] = \frac{ \frac{1}{2}^{n+1} }{\sqrt{n}} \rightarrow 0$$

as $n \rightarrow \infty$

Conclusion: the limit is 1.

 

[Billy Bob]

x=0?

$$B_n(x)=\sum_{k=0}^n \binom{n}{k} x^k (1-x)^{n-k} f(\frac{k}{n})$$

so $$B_n(0)=\sum_{k=0}^n \binom{n}{k} 0^k 1^{n-k} f(\frac{k}{n})$$

so $$B_n(0)=\binom{n}{0} 0^0 f(0)$$, which can't be helpful.

Also, $$B_n(1)=\sum_{k=0}^n \binom{n}{k} 1^k 0^{n-k} f(\frac{k}{n})$$

so $$B_n(1)=\sum_{k=0}^n \binom{n}{k} 1^k 0^{n-k} f(\frac{k}{n})$$

so $$B_n(1)=\binom{n}{n} 0^0 f(1)$$ which may not be 0 as I claimed earlier, but it still can't be helpful.

Oh well, average the two...

 

[dirk_mec1]

Oops

I see, if you take x =1/2 it matches my term exactly:
$$B_n \left( \frac{1}{2} \right) =\sum_{k=0}^n \binom{n}{k} \left( \frac{1}{2} \right) ^n f \left(\frac{k}{n} \right)$$Now

$$| B_n \left( \frac{1}{2} \right) - f \left(\frac{1}{2} \right)| \leq \frac{1}{2 \sqrt{n} }$$ $$\frac{1}{2 \sqrt{n}} + \ln \left( \frac{3}{2} \right) \longrightarrow \ln \left( \frac{3}{2}\ \right) \mbox{as}\ n \rightarrow \infty$$So the limit is 3/2!

 

[Billy Bob]

So the limit is 3/2!

Hooray! Isn't that nice!

I wrote $$\ln \left( \frac{3}{2} \right)-\frac{1}{2 \sqrt{n} }\le B_n \left( \frac{1}{2} \right) \le \ln \left( \frac{3}{2} \right)+ \frac{1}{2 \sqrt{n} }$$ to make the final steps crystal clear.

Don't forget to verify $$|f(x)-f(y)|\le|x-y|$$ so that the hint applies.

 

[dirk_mec1]

Hooray! Isn't that nice!

Yes it is :)

I wrote $$\ln \left( \frac{3}{2} \right)-\frac{1}{2 \sqrt{n} }\le B_n \left( \frac{1}{2} \right) \le \ln \left( \frac{3}{2} \right)+ \frac{1}{2 \sqrt{n} }$$ to make the final steps crystal clear.

Right, so the squeeze theorem applies.

Don't forget to verify $$|f(x)-f(y)|\le|x-y|$$ so that the hint applies.

I'm currently thinking on the proof for the hint but CS (after triangle inequality) gives me some nasty squares...