Solving Complex Math Problems: Recurrence Relations and Limits Explained

[hamsterman]

I'm trying to teach myself maths, and I have a couple of problems I can't figure out.

1. $\lim\sum\limits_{k=1}^{n}\frac{2k-1}{2^k}$. I see that this is $=\lim\sum\limits_{k=1}^{n}\frac{k}{2^{k-1}}-1$ but I have no idea what to do with that. I tried writing it as $\lim\frac{\sum\limits_{k=1}^{n}2^{n-k}k}{2^{n-1}}-1$ and then writing the top as $x_n$ where $x_1=1, x_k=2x_{k-1}+k$, but I failed. I hardly have any experience with recurrence relations.. And it seems overly complex.

2. $\lim\prod\limits_{j=1}^{n}(1+\frac{aj}{n^2}), a\in\mathbb{R}$. (It might be that a can only be positive, I don't remember..). I have no ideas about this at all. The answer is $e^{a/2}$, I think.
A related question, is $\lim\limits_{n\rightarrow\infty}(1+\frac{x}{n})^n=e^x$ ? I don't see how that works..

Thanks for your time

 

[susskind_leon]

To part 1: You apparently know the geometric series, right? If you express it in terms of
$$\sum_{n=0}^\infty r^n$$ and then look at the derivative with respect to r, what do you get?

 

[susskind_leon]

To part 2: If I were you, I would expand the product into a power series in a and see if you can find the right coefficients. It might be a tedious task though.

 

[hamsterman]

Thanks for the first. It seems that I've solved the second:
$\lim\prod\limits_{j=1}^{n}(1+\frac{aj}{n^2})=\lim\\sqrt[n^2]{\prod\limits_{j=1}^{n}(1+\frac{aj}{n^2})^{n^2}}=\lim\sqrt[n^2]{\prod\limits_{j=1}^{n}e^{aj}}=\lim e^{\frac{a}{n^2}\sum\limits_{j=1}^{n}j}=\lim e^{\frac{a}{n^2}\frac{n(n+1)}{2}}=e^{a/2}$

 

[susskind_leon]

Wow, that's what I call elegant :-o
Well done!