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I need to show that $\sum_{k=1}^n \frac{1}{k} - \ln n$ is decreasing. Can someone help please?

Clement C.
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Doe
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2 Answers2

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Set $$ u_n:=\sum_{k=1}^n \frac1k - \log(n), $$ then, $$ u_n-u_{n+1}=\log\left(1+\frac1n\right)-\frac1{n+1}, \quad n=1,2,3,\cdots. $$ Now observe that $$ \log\left(1+x\right)-\frac{x}{1+x}\geq0, \quad x \in [0,1].\tag1 $$ To see $(1)$, just consider the derivative, $$ \left(\log\left(1+x\right)-\frac{x}{1+x}\right)'=\frac{x}{(1+x)^2}\geq0,\quad x \in [0,1] $$ and the fact that $$\left(\log\left(1+x\right)-\frac{x}{1+x}\right)_{x=0}=0.$$

Then apply $(1)$ with $x=\dfrac1n$ to obtain $u_n-u_{n+1}\geq0$.

Olivier Oloa
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$$\left(\sum_{k=1}^{n+1} \frac1k - \ln(n+1)\right) - \left( \sum_{k=1}^{n} \frac1k - \ln n \right) = \frac{1}{n+1} + \ln\frac{n}{n+1}$$ but $$ \ln\frac{n}{n+1} = \ln\left(1-\frac{1}{n+1}\right) \leq -\frac{1}{n+1} $$ since $\ln(1+x) \leq x$.

For the last inequality, if it is not known: easy to show by convexity. Consider $f\colon x\mapsto \ln(1+x)$, which has $f(0)=0$, $f^\prime(0) = 1$, and is concave, so that $f(x) \leq f(0) + f^\prime(0)x$ for all $x \geq -1$).

Clement C.
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  • Or use the series for $\ln (1 - x)$ – vonbrand Jan 11 '16 at 22:43
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    Sure -- but that presupposes either a bit more knowledge, or a bit more faith in authority, than the simpler inequality above require. For reference: $$ \ln(1-x) = -\sum_{k=1}^\infty \frac{x^k}{k} = -x - \sum_{k=2}^\infty \frac{x^k}{k} $$ so for $x \geq 0$ we get $$ \ln(1-x) \leq -x. $$ – Clement C. Jan 11 '16 at 22:45