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How do I find out if $\sum\limits_n\log(1+{1\over n})$ diverges or converges? Wolfram recommends me to use comparison test, but I do not know series which diverges and less than this.

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    $\begingroup$ One can use Limit Comparison with $\sum \frac{1}{n}$. $\endgroup$ – André Nicolas Jan 3 '15 at 15:37
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Wolfram recommends me to use (some) compar(is)on test, but I do not know (any) series which diverges and (is) less than this (one).

$$\log\left(1+\frac1n\right)\geqslant\frac1{2n}$$

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$$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\log\prod_{n=1}^{N}\frac{n+1}{n}=\log(N+1).$$

If you want to use a comparison, notice that since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have: $$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\geq\frac{1}{n+1/2}$$ by Jensen's inequality, hence: $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)\geq 2\sum_{n=1}^{N}\frac{1}{2n+1} = 2H_{2n+1}-H_n.$$

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Since

$$\sum_{n = 1}^N \log\left(1 + \frac{1}{n}\right) = \sum_{n = 1}^N \log\left(\frac{n+1}{n}\right) = \sum_{n = 1}^N [\log(n+1) - \log(n)] = \log(N+1) \to \infty$$

the series $\sum_{n = 1}^\infty \log\left(1 + \frac{1}{n}\right)$ diverges.

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note $$\ln{\left(1+\dfrac{1}{n}\right)}=\dfrac{1}{n}+o(1/n)$$ since $$\sum_{n=1}^{\infty}\dfrac{1}{n}$$ is diverges

so $$\sum_{n=1}^{\infty}\ln{\left(1+\dfrac{1}{n}\right)}$$ is also diverges

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  • $\begingroup$ First statement isn't obvious to me. $\endgroup$ – Dark Archon Jan 3 '15 at 16:10
  • $\begingroup$ do you know $\ln{(1+x)}=x+o(x)?$ $\endgroup$ – math110 Jan 3 '15 at 16:11
  • $\begingroup$ What is $o(x)$? $\endgroup$ – Dark Archon Jan 3 '15 at 16:13
  • $\begingroup$ $x=\dfrac{1}{n}$ when $n\to +\infty$,then $x\to 0$ $\endgroup$ – math110 Jan 3 '15 at 16:25
  • $\begingroup$ @DarkArchon $f(x)=o(x)$ means that $\lim_{x\to 0}f(x)/x=0$. $\endgroup$ – Andrés E. Caicedo Jan 4 '15 at 0:52
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The first natural idea to understand how $\log(1+1/n)$ behaves when $n$ is small is to do Taylor expansion around $x=0$ for $\log(1+x)$, so that you have estimates for $\log(1+x)$. Then you get $$ \log(1+x) = \sum_{n \ge 1} \frac{(-1)^{n+1}}{n}x^n = x - x^2/2 + x^3/3 - \cdots, $$ by integrating the geometric series (and switching a minus sign) and by Taylor's theorem you get that there exists $1 \le \zeta \le x$ such that $$ \log(1+x) = x-\frac{\zeta^2}2 \ge x - \frac {x^2}2 = \frac{2x-x^2}2 = \frac{x(2-x)}2 \ge \frac x2 $$ for $0 \le x \le 1$ (because $\frac{x(1-x)}2 \ge 0$ on this interval).

A variant of this idea is to check that $\log$ is a concave function (the derivatives are $\frac 1x$ and $\frac {-1}{x^2}$), hence we can use Jensen's inequality : for all $\lambda \in [0,1]$, $$ \log(\lambda x + (1-\lambda) y) \ge \lambda \log(x) + (1-\lambda) \log(y) $$ so that for $y=1$ and $x=2$, we get $$ \log(1+ \lambda) \ge \lambda \log(2). $$ In both cases, we established an inequality of the form $\log(1+x) \ge cx$ for $x \in [0,1]$, which means $\log(1+1/n) \ge \frac cn$, hence your series diverges.

Hope that helps,

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Another way to think of this problem is that $\sum log(1+1/n)$ = $\log \prod(1+1/n)$ = $\log \prod((n+1)/n)$, which goes to infinity. Thus the sum diverges.

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  • $\begingroup$ Already explained on the page, twice. $\endgroup$ – Did May 11 '15 at 5:21

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