Euler Mascheroni finite sequence 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?
 A: 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 $\gamma_{n+1}<\gamma_n$.
A: 
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!
