User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)


 * $u \ v \ \mathsf{u} \ \mathsf{v} \ \nu \ \upsilon$

Anyone else have a hard time distinguishing between $u$ and $v$? I would like it to look more like this, does it confuse anyone else? It seems PW doesn't have the upgreek package. --GFauxPas 07:49, 27 January 2012 (EST)


 * Nope. Multiple years of extensive TeX writing and reading have trained my eye. I agree that referenced $v$ looks more distinguished, but imagine it is hard to implement. --Lord_Farin 08:08, 27 January 2012 (EST)

Convergence

 * I'm sorry I'm missing something. What do you mean a different sequence for each $n$?


 * What is the proper way to analyze $\displaystyle \left \vert{ \frac{n+1}{n-k+1}\frac{n^k}{ \left({n+1}\right)^k} \frac{ {a_{n+1} }^{k-1} } { {a_n}^{k-1} } }\right \vert$? --GFauxPas 11:31, 5 March 2012 (EST)


 * The critical point is that the sequence $\displaystyle \sum_{k=1}^n \binom n k \frac{a_n^{k-1}}{n^k}$ is not a sequence of partial sums. It is therefore void to try and assess its convergence by the ratio test. However, consider:


 * $\binom n k n^{-k} \le 1$ for $n,k\ge 1$ so $\displaystyle \sum_{k=1}^n \binom nk \frac{a_n^{k-1}}{n^k} \le \sum_{k=1}^n a_n^{k-1} = \frac{1-a_n^n}{1-a_n} \le \frac2{1-a_n}$ for $n$ large enough s.t. $a_n < 1$.


 * The latter expression converges to $2$ by continuity of $x\mapsto\dfrac2{1-x}$ and we are done. --Lord_Farin 12:49, 5 March 2012 (EST)


 * This proof works for real $a_n$ and so passing to the absolute value in case $a_n \in \C$ the result will be obtained as well. --Lord_Farin 12:50, 5 March 2012 (EST)


 * Why isn't it a series of partial sums? We have $\Sigma$ is running from $k=1$ to $n$ as $n \to \infty$, isn't that the same as $\displaystyle \sum_{k=1}^{\infty}$? Or is it because it needs to be o.t.f. $\displaystyle \sum_{k=1}^{\infty}b_k$ instead of $\displaystyle \sum_{k=1}^{\infty}b_n$, is that the problem? :( --GFauxPas 13:16, 5 March 2012 (EST)


 * The latter. The terms in the sum change as the upper limit increases; hence the difference of two consecutive terms in the partial sum sequence is not just the term with the highest index. This prevents it from being a conventional sequence of partial sums, so the theory for those does not apply. --Lord_Farin 13:24, 5 March 2012 (EST)


 * Okay, I'm making progress, thank you for your patience. But I'm still confused. How did you switch $n$ and $k$ in:
 * $\displaystyle \sum_{k=1}^n \binom nk \frac{a_n^{k-1}}{n^k} \le \sum_{n=1}^k a_n^{k-1}$? --GFauxPas 13:36, 5 March 2012 (EST)


 * Sorry, that's a (now corrected) typo. --Lord_Farin 13:41, 5 March 2012 (EST)


 * Okay, thanks. Another question, does Absolutely Convergent Series is Convergent work for $a_n \in \C$? Because it seems like we need it here. --GFauxPas 13:44, 5 March 2012 (EST)


 * Any $\R^n$ with Euclidean metric/norm is a Banach space. In particular, in this context, $\C$ with modulus is exactly the same as the Euclidean plane $\R^2$. That is, yes :). --Lord_Farin 13:50, 5 March 2012 (EST)

We have that $\displaystyle \lim_{n \to +\infty}a_n = 0$, by hypothesis.

To show that $\displaystyle \sum_{k=1}^n{n \choose k}\frac {{a_n}^{k-1}} {n^k}$ converges, observe that, for $n$ large enough:

am I on the right track? --GFauxPas 14:10, 5 March 2012 (EST)