The floor function is confusing me a bit.
Solve the series $\sum\limits_{ n=\lfloor\frac{m}{2} \rfloor}^m \frac1{n}$
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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
3 Answers
Hint:
If $n=2k$ is even
$\sum_{ n=\lfloor\frac{m}{2} \rfloor}^{m} \frac{1}{n}=\sum_{ n=k}^{2k} \frac{1}{n}=\sum_{ j=0}^{k} \frac{1}{k+j}=\frac{1}{k}\sum_{ j=0}^{k} \frac{1}{1+\frac{j}{k}}$
which is a Riemann sum. Can you calculate the corresponding integral?
The case $n$ is odd is either similar, or you can use the fact that you only get an extra term which converges to $0$...
Denote your expression by $f_m$, that is $ f_m = \sum\limits_{n = \left\lfloor {m/2} \right\rfloor }^m {\frac{1}{n}} . $ If $m$ is even, then $ \sum\limits_{n = m/2}^m {\frac{1}{n}} , $ and, as you can easily verify, $ f_{m+1} = f_m + \frac{1}{{m + 1}}. $ Hence the case $m$ odd follows immediately from the case $m$ even: $f_3 = f_2 + 1/3$, $f_5 = f_4 + 1/5$, $f_7 = f_6 + 1/7$, etc.
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0So $f_m$ is...? – 2011-09-10
Hint. m even $\implies n = \frac{m}{2} = \frac{2k}{2} = k$, m odd $\implies n = \frac{m-1}{2}$ since $\frac{m}{2} = \frac{2k+1}{2} = k + \frac{1}{2}$ for some k.
Edit:
Let $H(m) = \sum_{n=1}^{m} \frac{1}{n}$. For even m = 2k, then $\sum_{ n=\lfloor\frac{m}{2} \rfloor}^{m} \frac{1}{n} = H(2k) - H(k)$.
Now H(2k) - H(k) is equivalent to the alternating sum = $1 - \frac{1}{2} + \frac{1}{3} - \dots - \frac{1}{2k}$, and so approaches ln(2). This can be proved using induction.
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0I edited my post, please check. – 2011-07-12