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
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:

Recall that $a_n \to 0$, by hypothesis.

This means that $0 \le \left\vert{a_n}\right\vert \le 1$ for sufficiently large $n$, because $a_n$ is Cauchy.

Then, we can say:


 * $\displaystyle \left \vert{ \frac{ 1 - {a_n }^k } {1 - a_n} }\right\vert \le \left \vert{ \frac{ 1 - a_n } {1 - a_n} }\right\vert = 1$ as $n \to +\infty$, because $k \ge 1$.

Hence $\displaystyle \left \vert{\sum_{k=1}^n{n \choose k}\frac { {a_n}^{k-1} } {n^k} }\right\vert$ converges, by the Comparison Test.

That means that $\displaystyle \sum_{k=1}^n{n \choose k}\frac { {a_n}^{k-1} } {n^k}$ converges as well, because Absolutely Convergent Series is Convergent.

Is this okay? --GFauxPas 08:54, 6 March 2012 (EST)


 * Is it legitimate to think of it this way? The lower index is being "pulled down" one, and so in addition to replacing all indices with the new "shifted" index, the upper index is "pulled down" the same degree as the lower index? That's how my brain is interpreting it... --GFauxPas 18:21, 5 March 2012 (EST)


 * That's precisely it. --Lord_Farin 02:17, 6 March 2012 (EST)


 * There's a much-linked-to but as-yet-unwritten page Permutation of Indices‏ which proves this. It is explained in detail in an early chapter of, but I never got round to posting it up. --prime mover 02:49, 6 March 2012 (EST)