Calculating with summations

$$b! \left (\sum_{n=0}^{\infty}\frac{1}{n!}-\sum_{n=0}^{b}\frac{1}{n!} \right )=\sum_{n=b+1}^{\infty}\frac{b!}{n!}> 0$$

I don't understand why $n=b+1$ in the last step of the expression.

• Can explain a bit more where this all comes from? What do all the constants signify? Any restraints? Nice formatting, though. Commented Jan 7, 2017 at 19:58
• It comes from Fourier's proof, en.m.wikipedia.org/wiki/Proof_that_e_is_irrational Commented Jan 7, 2017 at 20:00
• @TheCount if you liked the formatting, look at this link: codecogs.com/latex/eqneditor.php . Just remember to use the dollar signs at the end and start of the equation - I'm new to this and made the same mistake myself. Commented Jan 7, 2017 at 20:16
• Oh I know how to do it, I was just complimenting you. So many new users never use LaTeX and it drives me nuts. Commented Jan 7, 2017 at 20:27
• @TheCount Oh that I did notice when I posted my first question. Commented Jan 7, 2017 at 20:34

2 Answers

The terms for $n=0$, $n=1$, ..., $n=b$ are subtracted, so they should not be part of the final summation.

• Now I understand. Commented Jan 14, 2017 at 15:09
• @theHumbleOne thank you. Commented Jan 14, 2017 at 15:12

If the series $\sum_{n=0}^\infty a_n$ converges then the series $\sum_{n=r}^\infty a_n$ also converges for any natural $r$ and $$\lim_{k\to\infty}\left(\sum_{n=0}^ka_n-\sum_{n=0}^ba_n\right)=\lim_{k\to\infty}\sum_{n=b+1}^k a_n=\sum_{n=b+1}^\infty a_n$$