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prove that the sequence converges $$a_n = \sum_{k=1}^n \frac1k \ -\ \int_1^n\frac{dx}x$$

Attempt:

I dont even know where to start here. the integral gives you -ln(n)

sarah
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  • I'm not sure what you're asking. Please use latex. – Suugaku Dec 16 '13 at 00:13
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    Monotonic and bounded – Berci Dec 16 '13 at 00:14
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    The integral of second term is not $-\frac{1}{n^{2}}+1$. It is $-\ln(n)$ – fiverules Dec 16 '13 at 00:18
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    For the convergence, observe that 1. The sequence is DECREASING. 2. The sequence is BOUNDED BELOW by 0 – fiverules Dec 16 '13 at 00:21
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    Hint: google "Euler_Mascheroni constant", or "gamma constant" – DonAntonio Dec 16 '13 at 00:25
  • @sarah : this question looks identical to http://math.stackexchange.com/questions/607039/how-to-show-gamma-aka-eulers-constant-is-convergent . One of the answers to the old question has a nice picture (unlike the first two answers to appear here). However, I prefer ncmathsadist's answer below to the old answers because it does not require asymptotic notation, which most calculus students don't know. I have not read Ethan's answer below. – Stefan Smith Dec 16 '13 at 01:42

2 Answers2

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$$a_{n+1} - a_n = \sum_{k=1}^{n+1}{1\over k} - \sum_{k=1}^{n}{1\over k} - \int_n^{n+1} {dt\over t} = {1\over n+1} - \int_n^{n+1}{dt\over t} = \int_n^{n+1}\left({1\over n+1} - {1\over t}\right) dt < 0$$ The sequence decreases, and it is clearly bounded below by 0.

ncmathsadist
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$$\frac{1}{\lfloor{x}\rfloor}=\frac{1}{x}+\frac{ \{ x \}}{x\lfloor{x}\rfloor}$$

$$\ln(n)+\int_{1}^n\frac{ \{ x \}}{x\lfloor{x}\rfloor}dx=\int_{1}^n\frac{1}{\lfloor{x}\rfloor} dx=\sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{1}{\lfloor{x}\rfloor} dx=\sum_{k=1}^{n-1}\frac{1}{k}$$

$$a_n=\frac{1}{n}+\int_{1}^n\frac{ \{ x \}}{x\lfloor{x}\rfloor}dx=O(1)$$

Ethan Splaver
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