Define $\gamma=\lim_{n\to\infty} \sum_{k=1}^n 1/k - ln(n)$.
I know that $\gamma$ is nonnegative, but i don't know how to prove that it is positive.
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Each "term" $\frac{1}{k}-\ln k$ of the sum is positive. – André Nicolas Mar 30 '14 at 04:32
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@Andre How does that imply that $\gamma$ is positive? ${1/n}$ converges to 0 even though each term is positive – John. p Mar 30 '14 at 04:36
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We have a sum of positive terms, so the sum is increasing. – André Nicolas Mar 30 '14 at 04:38
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@Andre ${\sum_{k=1}^n 1/k - \log n}$ is a decreasing sequence – John. p Mar 30 '14 at 04:47
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see http://math.stackexchange.com/questions/306371/simple-proof-of-showing-the-harmonic-number-h-n-theta-log-n/306379#306379 – Will Jagy Mar 30 '14 at 04:51
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@John.p: Yes, I must have been thinking of $\sum_1^n \frac{1}{k}-log(n+1)$, same limit. – André Nicolas Mar 30 '14 at 05:03
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A simple comparison suffices: consider the functions $$f(x) = \frac{1}{x}, \quad g(x) = \frac{1}{\lfloor x \rfloor}.$$ Clearly, $g(x) \ge f(x)$ for all $x \ge 1$. Integrating both on $[1, n]$ gives $$\int_{x=1}^n f(x) \, dx \le \int_{x=1}^n g(x) \, dx,$$ or $$\log n \le \sum_{k=1}^{n-1} \frac{1}{k} < \sum_{k=1}^n \frac{1}{k}.$$
heropup
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I know that $\sum_{k=1}^n \frac{1}{k} - \log n >1/n$. But how does that gurantee that its limit is positive – John. p Mar 30 '14 at 04:46
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Think about the picture: Consider $$D_k = \int_{x=k}^{k+1} \frac{1}{\lfloor x \rfloor} - \frac{1}{x} , dx = \frac{1}{k} - \log(k+1) + \log k > 0.$$ Then the limit of the partial sums of $S_n = \sum_{k=1}^n \frac{1}{k} - \log (n+1)$ is simply $$S_\infty = \sum_{k=1}^\infty D_k > 0,$$ and your sum is even larger than that. – heropup Mar 30 '14 at 04:51
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see http://math.stackexchange.com/questions/306371/simple-proof-of-showing-the-harmonic-number-h-n-theta-log-n/306379#306379 – Will Jagy Mar 30 '14 at 04:52