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$\lim _{n\to \infty }\left(\frac{1+5+5^2+...+5^{n-1}}{1-25^n}\right)$

I have a solution to this question, but I don't really understand it. It's:

$\lim _{n\to \infty \:}\left(\frac{1+5+5^2+\ldots \:+5^{n-1}}{1-25^n}\right)=\:\lim _{n\to \infty }\left(\frac{1-5^n}{\frac{1-5}{1-25^n}}\right)=\:-\frac{1}{4}\lim _{n\to \infty }\left(\frac{1}{1+5^n}\right)=0$

Thanks in advance.

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up vote 4 down vote accepted

The first step follows from computing the (finite) geometric sum explicitly: We have

$$1 + x + x^2 + ... + x^{n - 1} = \frac{1 - x^n}{1 - x}$$

The second step is algebra.

The third step factors $1 - 25^n$ as a difference of squares and cancels one term:

$$1 - 25^n = (1 - 5^n)(1 + 5^n)$$

The final step uses the fact that $5^n \to \infty$ as $n$ grows, and hence the limit is zero.

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Thank you very much! – user165204 Jul 20 '14 at 8:34

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