Let
$\gamma_n = \sum_{k=1}^n\frac1k - \ln(n) $ , for n in the natural numbers.
How do you show that $\gamma_{n+1} < \gamma_n$ using Taylor's theorem?
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Does this question necessitate my use of Taylor's theorem? Or am I allowed to be... creative? – Simply Beautiful Art Mar 03 '17 at 00:16
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the objective is to use Taylor's thm, otherwise there are easier/simpler ways to prove the statement. – bashmike Mar 03 '17 at 00:23
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I'm almost certain that the inequality sign should be flipped. The sequence from which the euler-mascheroni constant is derived is known to be decreasing, not increasing. – Hrhm Mar 03 '17 at 01:30
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You're absolutely right, my mistake – bashmike Mar 03 '17 at 01:31
2 Answers
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With Taylor:
$$\begin{align}\gamma_{n+1}&=\sum_{k=1}^{n+1}\frac1k-\ln(n+1)\\&=\sum_{k=1}^n\frac1k-\ln(n)+\ln(n)+\frac1{n+1}-\ln(n+1)\\&=\gamma_n+\frac1{n+1}-\ln\left(\frac{n+1}n\right)\\&=\gamma_n+\frac1{n+1}-\ln\left(1+\frac1n\right)\\&=\gamma_n+\frac1{n+1}-\sum_{k=1}^\infty\frac{(-1)^{k+1}}{kn^k}\\&<\gamma_n+\frac1{n+1}-\frac1n\\&<\gamma_n\end{align}$$
Where we note that
$$\underbrace{\left(\frac1n-\frac1{2n^2}\right)}_{<\frac1{n^3}}+\underbrace{\left(\frac1{3n^3}-\frac1{4n^4}\right)+\dots}_{<0}<\frac1n$$
For all $n>1$. Cheching $n=1$ separately is not hard.
Without Taylor:
$$\begin{align}\gamma_{n+1}&=\sum_{k=1}^{n+1}\frac1k-\ln(n+1)\\&=\sum_{k=1}^n\frac1k-\ln(n)+\ln(n)+\frac1{n+1}-\ln(n+1)\\&=\gamma_n+\frac1{n+1}+\ln(n)-\ln(n+1)\\&=\gamma_n+\frac1{n+1}+\int_1^n\frac1x\ dx-\int_1^{n+1}\frac1x\ dx\\&=\gamma_n+\frac1{n+1}-\int_n^{n+1}\frac1x\ dx\\&<\gamma_n+\frac1{n+1}-\int_n^{n+1}\frac1{n+1}\ dx\\&=\gamma_n+\frac1{n+1}-\frac1{n+1}\\&<\gamma_n\end{align}$$
Thus, it is elementarily proven that $\gamma_{n+1}<\gamma_n$.
Simply Beautiful Art
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1Perfect answer. I would complete a bit with this lemma. Let $\sum (-1)^ka_k$ and alternating series such that $(a_k)$ is positive and monotone then
$$\left|\sum_{k=0}^\infty(-1)^k a_k-\sum_{k=0}^n(-1)^k a_k\right|\le a_{k+1}$$
– Masacroso Mar 03 '17 at 00:39 -
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I dont know if it have a name but it can be proved writing the series as the addition of two series of different signs. – Masacroso Mar 03 '17 at 00:48
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Sorry what I meant to say is that $$\int_{n}^{n+1}\frac{1}{x},dx>\int_{n}{n+1}\frac{1}{n+1}$$ Since for $x\in[n,n+1]$, $$\frac{1}{x}>\frac{1}{n+1}$$ Taking into account the negative signs, the inequality should be flipped. – Hrhm Mar 03 '17 at 01:41
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Sorry, I don't mean to be rude, but I think this new proof must be wrong as well. If we follow your previous second proof verbatim, only replacing the last step by switching the inequality sign, then we have clearly shown that $y_{n+1}<y_n$. We can also show that your first proof is fallacious by looking at the second portion you have highlighted in the Taylor expansion of $\ln\left(1+\frac{1}{n}\right)$, which is actually positive for $x>1$. For further confirmation that the series is decreasing: http://math.stackexchange.com/questions/629630/simple-proof-euler-mascheroni-gamma-constant – Hrhm Mar 03 '17 at 01:51
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@Hrhm ah, you are right, I have all the inequalities backwards :-/ Thanks for the proof-reading! – Simply Beautiful Art Mar 03 '17 at 01:59
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I thought it might be instructive to present an approach that relies on only a standard inequality that can be obtained with pre-calculus tools. To that end, we begin with a short primer.
PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1 }\tag 1$$
Let $\gamma_n=\sum_{k=1}^n\frac1k-\log(n)$. Then,
$$\gamma_{n+1}-\gamma_n=\frac{1}{n+1}-\log\left(1+\frac1n\right) \tag 2$$
Setting $x=1+\frac1n$ in the the left-hand side inequality in $(1)$ and applying it to $(2)$ reveals
$$\gamma_{n+1}-\gamma_n\le \frac{1}{n+1}-\frac1{n+1}=0$$
Therefore, $\gamma_{n+1}\le \gamma_n$ and we are done!
Mark Viola
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