I was struggling to find the sum of $\sum_{i=n/2}^n{1/i}$. I came up with a series $$S = \frac{2}{n} + \frac{2}{(n+2)}+\frac2{(n+4)}+\ldots+\frac{2}{2n}$$ But I am not being able to find which formula to use to find thi sum
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1You either write it as $H_{n} - H_{n/2-1}$ or not at all. – Kenny Lau Sep 09 '17 at 16:52
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1Is this the full problem as is delivered unto you? – Kenny Lau Sep 09 '17 at 16:52
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2There is no simple closed formula for Harmonic Numbers and your expression is just a difference of two of these. – lulu Sep 09 '17 at 16:52
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A compendium of techniques of studying the limit of these sums has been collected here. – Jyrki Lahtonen Sep 09 '17 at 18:55
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There's not a nice answer, but you can make a good approximation with
$$\sum_{i=n/2}^n \frac{1}{i} \approx \int_{n/2-1}^{n} \frac{1}{x} \; dx = \ln \frac{2n}{n-2}.$$
There's a little problem that $n/2$ might not be an integer, so the above assumes $n$ is even. If $n$ is odd, change the lower integration limit to $(n-1)/2.$
