Problems with limits

  • Thread starter hamsterman
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  • #1
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I'm trying to teach myself maths, and I have a couple of problems I can't figure out.

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

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

Thanks for your time
 

Answers and Replies

  • #2
To part 1: You apparently know the geometric series, right? If you express it in terms of
[tex] \sum_{n=0}^\infty r^n [/tex] and then look at the derivative with respect to r, what do you get?
 
  • #3
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.
 
  • #4
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Thanks for the first. It seems that I've solved the second:
[itex]\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}[/itex]
 
  • #5
Wow, that's what I call elegant :-o
Well done!
 

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