Thread: Problem 39
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Old October 31st, 2007, 10:26 AM
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Maybe I'm off base here, PH,:

\lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\frac{1}{na+b}\right)=\frac{1}{a}

\lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\frac{1}{nc+d}\right)=\frac{1}{c}

So, we have: \frac{\frac{1}{a}}{\frac{1}{c}}=\boxed{\frac{c}{a}}