Proving monotonicity and convergence of a sequence $a_n=(1+{1 \over 2}+{1 \over 3}+....+{1 \over n})-\ln(n)$
Show that $a_n$ is bounded and monotone and hence convergent. 
I know that the $-\ln(n)$ portion will be monotonically decreasing. I think I need to somehow show that $1+{1 \over 2}+{1 \over 3}+....+{1 \over n}>\ln(n)$ so that the whole sequence will be monotonically increasing ie. $a_n<a_{n+1}$ and also need to show that for all $n$ there is a $C$ such that $a_n<C$ for all $n$
 A: I am going to prove that it should be bounded. It's convenient to write $\ln{x}$ as $\int_{1}^{n} \frac{1}{x} dx$. It's easy to confirm that $\frac{1}{x}$ is decreasing function. So, we have that :
$$ \frac{1}{i+1} \le \int_{i}^{i+1} \frac{1}{x} \,dx \le \frac{1}{i} $$
By taking sigma, we have that :
$$\sum_{i = 1}^{n-1} \frac{1}{i+1} \le \sum_{i = 1}^{n-1} \int_{i}^{i+1} \frac{1}{x} \,dx \le \sum_{i = 1}^{n-1} \frac{1}{i} \implies $$
$$ \sum_{i = 1}^n \frac{1}{i} - 1 \le \int_{1}^{n} \frac{1}{x} \,dx \le \sum_{i = 1}^n \frac{1}{i} - \frac{1}{n} \implies $$
$$ \frac{1}{n} \le \sum_{i = 1}^n \frac{1}{i} - \int_{1}^{n} \frac{1}{x} \,dx \le 1 $$
So, it's true that it's bounded. 
For the monotonity, previously we have that $$ \frac{1}{n+1} \le \int_{n}^{n+1} \frac{1}{x} \, dx$$
So, if we assume $$F(n) = \sum_{i = 1}^n \frac{1}{i} - \int_{1}^{n} \frac{1}{x}$$. We will have that :
$$F(n+1) - F(n) = \frac{1}{n+1} - \int_{n}^{n+1} \frac{1}{x} \le 0$$. SO, it's true that $F(n)$ monotonically decreasing and bounded $\square$ 
A: Same idea written differently:
$H_n -1 \le \ln n \le H_n$, hence $|a_n| \le 1$.
Since $\ln$ is concave, we have $\ln (1+x) \le x$ for $x >0$.
Hence $\ln n - \ln (n+1) = \ln({n \over n + 1})=\ln(1-{1 \over n+1}) \le -{1 \over n+1}$, we see that
$\ln n - \ln (n+1) \le H_n - H_{n+1}$, or $a_{n+1} \le a_n$.
A: This answer may be off-topic; so, please, forgive me if this is the case.
$$\sum_{i=1}^n\frac 1i=H_n$$ the rhs being the harmonic number. For large values of $n$, the asymptotic expansion is $$H_n=\gamma +\log (n)+\frac{1}{2 n}-\frac{1}{12
   n^2}+O\left(\frac{1}{n^3}\right)$$ This makes $$a_n=\sum_{i=1}^n\frac 1i-\log(n)=\gamma +\frac{1}{2 n}-\frac{1}{12
   n^2}+O\left(\frac{1}{n^3}\right)$$
A: There is a nice "graphical proof" of the convergence of the (increasing) parent sequence
$$1-a_n=\ln(n)-\left({1 \over 2}+{1 \over 3}+....+{1 \over n}\right)$$
towards a limit: move the different involved surfaces into the unit square. As they do not overlap, the blue sawtooth region converges to a limit region whose area, slightly smaller than $1/2$, is $1-\gamma \approx 0.423...$, where $\gamma$ is Euler's constant.
 
