Prove that the sequence sum(i=1 --> n) 1/(10^i) converges.

I'm pretty sure it converges to 1/9 but how to we get there by fixing epsilon?

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- Sep 15th 2010, 09:30 PMDavyHilbertEpsilon Convergence Proof
Prove that the sequence sum(i=1 --> n) 1/(10^i) converges.

I'm pretty sure it converges to 1/9 but how to we get there by fixing epsilon? - Sep 15th 2010, 10:50 PMProve It
I'm not sure what you mean by "epsilon convergence" but this is a geometric series.

You should know that for a finite geometric series

$\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}$.

Therefore it is convergent.

You should also be able to see that if you make $\displaystyle n \to \infinity$, i.e. make this an infinite series, since $\displaystyle |r| < 1$, that means $\displaystyle r^n \to 0$ and thus the sum goes to $\displaystyle S_{\infty} = \frac{a}{1 - r}$, which is also convergent. - Sep 16th 2010, 08:14 AMDavyHilbert
We need to prove convergence by fixing ε > 0 and showing $\displaystyle absval(a_n - L) = absval((sum(i=1 ->n) 1/(10^i) - 1/9) < $ε

- Sep 16th 2010, 08:35 AMDefunkt
1) You can simply write | | instead of absval, even with LaTeX.

2) As Prove It said,

$\displaystyle \displaystyle \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{10^i} = \lim_{n \to \infty} \frac{1}{10} \cdot \frac{1 - 10^{-n}}{1 - 10^{-1}}$

Double click on the latex to see the correct code.