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Test the series for convergence or divergence
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\[\sum_{n=1}^{\infty}\frac{ 1 }{ 2^{n}-1 }\]
converges like there is not tomorrow
*no tomorrow compare it to the geometric series \[\sum\frac{1}{2^n}=1\]
i need to test it by the limit comparison test
yes because \(2^n>2^n-1\) so \[\frac{1}{2^n}<\frac{1}{2^n-1}\] so you do need the limit comparison test
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but since \[\lim_{n\to \infty}\frac{2^n}{2^n-1}=1\] you win right away
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