Solving Complex Math Problems: Recurrence Relations and Limits Explained

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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
 
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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?
 
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.
 
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}
 
Wow, that's what I call elegant :-o
Well done!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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