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Let $x_n = \sum_{k=1}^{n} \frac{1}{k} - \log n$. Prove that $x_n$ converges as $n \to\infty$.

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  • $\begingroup$ The (finite) limit $\gamma$ of the sequence $\{x_n\}$ is well-known: it is the Euler constant. Check here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant $\endgroup$ – Ángel Valencia Dec 1 '15 at 1:55
  • $\begingroup$ Do you know Stieltjes integration? $\endgroup$ – Clayton Dec 1 '15 at 1:55
  • $\begingroup$ @ÁngelValencia So if the limit is well known, why would this be false? $\endgroup$ – 5xum Dec 1 '15 at 1:56
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    $\begingroup$ Sorry, it was my mistake. I didn't read well. $\endgroup$ – Ángel Valencia Dec 1 '15 at 1:56
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    $\begingroup$ Try using left and right Riemann sums to bound $\sum_{k=1}^n \frac{1}{k}$ by something involving $\log (n)$ (since $\int_1^n \frac{1}{x}dx= \log(n)$). $\endgroup$ – kccu Dec 1 '15 at 2:01
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Hint: A proof without using the integration concepts.

  • Show that $$\frac{1}{n + 1} < \log\left(1 + \frac{1}{n}\right) < \frac{1}{n}. \tag{1}$$ You might need the facts that the sequence $\left(1 + \frac{1}{n}\right)^n$ increasingly converges to $e$ and the sequence $\left(1 + \frac{1}{n}\right)^{n + 1}$ decreasingly converges to $e$.
  • Use $(1)$, show that $$\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n + 1} < \log(n + 1) < 1 + \frac{1}{2} + \cdots + \frac{1}{n}. \tag{2}$$

  • Use $(2)$, show that the sequence $\{x_n'\}$ is increasing and bounded above, where $$x_n' = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \log (n + 1).$$ Hence $x_n'$ converges.

  • Show that $x_n$ and $x_n'$ have the same limit.

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  • $\begingroup$ Good outline. An alternative route is to show that $x_n' = \int_{1}^{n+1}\left[\frac{1}{\lfloor x\rfloor} - \frac{1}{x}\right]{\rm d}x$ and bound the integrand above to conclude that $x_n'$ converges. $\endgroup$ – Kibble Dec 1 '15 at 2:25
  • $\begingroup$ Well done ... +1. $\endgroup$ – Mark Viola Dec 1 '15 at 3:03

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