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If $|a_n| < 10^{-n}$, prove that $\sum^{\infty}_{n=1} a_n$ converges.

Could someone give me a hint as to how to start this?

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Hint If $|a_n|<10^{-n}$, then $\sum|a_n|<\sum 10^{-n} <\infty$

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To be even more sure : $\sum_{n\geq 1} 10^{-n} = 10^{-1}/(1-10)=1/90$, as a geometric series. – Student Apr 3 '12 at 5:03
To be sure even more, using the fact that, absolute convergence leads to convergence - we done. – Salech Alhasov Apr 3 '12 at 5:09
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If $A_n = \sum^{n}_{i=1} a_i$, if $n > m$, $$A_n - A_m = \sum^{n}_{i=m+1} a_i < \sum^{n}_{i=m+1} 10^{-i} < 10^{-m}$$ (you can do better, but this is enough). Then apply the Cauchy criterion.

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Then $|A_{n} - A_{m}| < 10^{-m}$. Since $10^{-m} \to 0$, $|A_{n} - A_{m}| < \epsilon$ for $n > m \geq N$. – user26139 Apr 3 '12 at 1:45

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