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If Euler's constant is described as the limit as n approaches infinity of the following:

$$t_n = 1 + \frac 12 + \frac 13 \cdots + \frac1n -\ln(n)$$

How can one prove that $t_n$ is greater than $0$ for all values of $n$?

Thanks!

corindo
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user270893
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5 Answers5

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First we have

$$\log (k+1)-\log k=\int_k^{k+1}\frac{1}{u}\,du\le \frac{1}{k} \tag 1$$

Next, we can sum both sides of $(1)$ to reveal that

$$\sum_{k=1}^n \left(\log (k+1)-\log k\right)=\log (n+1)\le \sum_{k=1}^n \frac{1}{k}\tag 2$$

Thus, we see that $(2)$ implies that

$$\sum_{k=1}^n \frac{1}{k}-\log (n+1)\ge 0$$

and therefore

$$\sum_{k=1}^n \frac{1}{k}-\log n > 0$$

Mark Viola
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  • Could you give more detail on how you made the jump between each step? Im not great at this subject at all.. Thank you so much! – user270893 Sep 15 '15 at 06:12
  • Sure. We used the definition of the log function $\log x=\int_1^x \frac{1}{u},du$ and the fact that $\frac{1}{u}\le \frac{1}{k}$ when $k\le u\le k+1$. We summed a telescoping sum $\sum_{k=1}^n(\log(k+1)-\log k)=(\log(n+1)-\log n)+(\log n-\log (n-1))+(\log (n-1)-\log (n-2)),\cdots +(\log (2)-\log (1))=\log (n+1)$. Finally, we used the fact that $\log (n+1)>\log(n)$ – Mark Viola Sep 15 '15 at 06:20
  • Ahh I see! Thank you!! – user270893 Sep 15 '15 at 06:24
  • Why the down vote?? Whoever gave this a down vote, please explain. – Mark Viola Sep 15 '15 at 06:24
  • You're welcome. My pleasure. – Mark Viola Sep 15 '15 at 07:20
  • (+1) I am often baffled by the downvotes to good answers. It is best not to dwell on them, and not to expect most downvoters to explain. Only rarely has a downvoter explained their vote, but when they do, I always appreciate it since it allows me to improve my answer. – robjohn Sep 15 '15 at 08:54
2

First, put $s_n = 1+\dfrac{1}{2}+\cdots + \dfrac{1}{n}$, then

$1+\dfrac{1}{2}+\cdots + \dfrac{1}{n} - \ln n > 0$ because:

$1 = \displaystyle \int_{1}^2 1dx \geq \displaystyle \int_{1}^2 \dfrac{1}{x}dx$,

$\dfrac{1}{2} = \displaystyle \int_{2}^3 \dfrac{1}{2}dx \geq \displaystyle \int_{2}^3 \dfrac{1}{x}dx$, and

$.....$

$\dfrac{1}{n-1} = \displaystyle \int_{n-1}^n \dfrac{1}{n-1}dx\geq \displaystyle \int_{n-1}^n \dfrac{1}{x}dx$, and add these integrals we have:

$s_n = s_{n-1} + \dfrac{1}{n} \geq \displaystyle \int_{1}^n \dfrac{1}{x}dx + \dfrac{1}{n}=\ln n + \dfrac{1}{n} > \ln n\Rightarrow t_n = s_n - \ln n > 0$

DeepSea
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By Abel's summation $$\sum_{k\leq n}\frac{1}{k}=1+\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t^{2}}dt\geq1+\int_{1}^{n}\frac{t-1}{t^{2}}dt=\log\left(n\right)+\frac{1}{n}>\log\left(n\right) $$ where $\left\lfloor t\right\rfloor $ is the floor function.

Marco Cantarini
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1

Since the function $x \mapsto 1/x: ]0, \infty[ \to \mathbb{R}$ is strictly decreasing, we have $\sum_{k=1}^{n}\frac{1}{k} > \int_{x=1}^{n+1}\frac{1}{x} = \log (n+1)$ for all $n \geq 1$. But, since $\log (n+1) > \log n$ for all $n \geq 1$, we have $\sum_{k=1}^{n}\frac{1}{k} - \log n > 0$ for all $n \geq 1$.

Yes
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Since $1+x\le e^x$ for $x\in\mathbb{R}$, we have that $\log(1+x)\le x$ for $x\gt-1$. Thus, for $x=-\frac1n$, $$ \begin{align} \frac1n-\log\left(\frac{n}{n-1}\right) &=\log\left(1-\frac1n\right)-\left(-\frac1n\right)\\ &\le0\tag{1} \end{align} $$ Inequality $(1)$ means that $$ \sum_{k=1}^n\frac1k-\log(n)\tag{2} $$ is a decreasing sequence.

However, using the standard comparison of sum and integral of a decreasing function, $$ \begin{align} \sum_{k=1}^n\frac1k &\ge\int_1^{n+1}\frac1x\,\mathrm{d}x\\ &=\log(n+1)\\[8pt] &\ge\log(n)\tag{3} \end{align} $$

Inequality $(3)$ answers the question. In addition, it shows that $(2)$ is not only a decreasing sequence, but also bounded below by $0$. This means that $(2)$ decreases to a non-negative limit.

This limit is known as $\gamma$, the Euler-Mascheroni Constant, and a fairly simple and efficient method for computing $\gamma$ is derived in this answer.

Community
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robjohn
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