The floor function is confusing me a bit.
Solve the series $\sum\limits_{ n=\lfloor\frac{m}{2} \rfloor}^m \frac1{n}$
2
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calculus
sequences-and-series
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0Could you elaborate on your confusion? Can you sum from n = 2 to 4 ? How about from n = 3 to 7 ? – 2011-07-12
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0I understand how the floor function works, I'm just confused about how to deal with it in this situation. I've expanded the the series looking for a general solution in terms of $m$, but I can't seem to find anything concise. – 2011-07-12
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1@textra: You will not find a pleasant explicit formula, because nobody will. What is the context? Is it a good approximation that is needed? (For largish $m$ the sum is about $\ln(2)$.) – 2011-07-12