Consider the Euler-Maschenori constant defined by \gamma=\lim_{n\to \infty}\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}-\ln n\right )=0.57721566\cdots . No one knows if this number is rational or irrational(so don't try it now).  However we are going to show that this limit is indeed a real number. Following is the proof in steps. Identify the lines containing only True statements.

Let u_n=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}-\ln n where n is a positive integer. 

(1) Define f(n)= u_n-u_{n-1}=\frac{1}{n}-\ln n+\ln (n-1)=\frac{1}{n}+\ln(1-\frac{1}{n})

(2) Consider the real valued function f(x)=\frac{1}{x}+\ln (1-\frac{1}{x}). Now f'(x)=\frac{1}{x^2(x-1)}>0 for x>1. So f(x) is increasing for x>1.

(3) Therefore f(n) is increasing for n\geq 2.

(4) Also we have \lim_{n\to\infty}f(n)=0. Therefore \sup\{f(n)|n\geq 2\}=0 and f(n)\leq 0 for n\geq 2. So u_n is decreasing for n\geq 2.

(5) Using an inequality we used in the proof of the Integral Test, we have \sum_{k=1}^n \frac{1}{k}>\int_1^{n+1}\frac{dx}{x}=\ln(1+n)>\ln n so u_n is bounded below by 0.

(6) Therefore u_n is converging and \lim_{n\to\infty}u_n is a real number.

(7) Note also that \sum_{n=1}^{\infty}\frac{1}{n} is diverging to \infty. Consider the partial sums S_m=\sum_{n=1}^m \frac{1}{n}. Using the inequalities mentioned in part 5., we can show that a value of m for which S_m\leq 100 is m=9\times10^{42}.