How do you show that the sequence $$a_n=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}\right)-\ln(n)$$ is convergent.
Give a convincing argument, that $a_n$ converges to a number called gamma. Using a picture if you want. Remember the anti-derivative of $\ln(x)$ is $\frac 1x$
I am not sure how to give a convincing argument that this sequence converges to $\gamma$ I never heard of this number gamma before.
$a_{n+1} - a_n = \dfrac{1}{n+1} - \ln(n+1)+\ln n = \dfrac{1}{n+1} - \ln\left(1+\dfrac{1}{n}\right)= \dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}-\ln\left(1+\dfrac{1}{n}\right)= \dfrac{x}{1+x} - \ln(1+x)=1-\dfrac{1}{x+1} -\ln(1+x)=f(x), 0< x = \dfrac{1}{n}<1\Rightarrow f'(x) = \dfrac{1}{(x+1)^2}-\dfrac{1}{x+1}<0\Rightarrow a_{n+1}-a_n = f(x) < f(0) = 0\Rightarrow a_{n+1} < a_n\Rightarrow \{a_n\} \text{ is a strictly decreasing sequence, and further it is bounded below by $0$ because } a_n = \dfrac{1}{n} + \left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n-1} - \ln n\right) > \dfrac{1}{n} + 0 = \dfrac{1}{n} > 0, \text{ thus it converges .}$
Hint: Try sketching the graph of 1/x and consider rectangles of area 1,1/2,1/3........ etc under the graph. Then use what you know about the integral of 1/x and also what you know about increasing bounded sequences. Hope this helps.
