For $n\ge 1$, define $a_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n$. Prove that $\lim_{n\to\infty} a_n$ exists. 
Question: For $n\ge 1$, define $a_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n$. Prove that $\lim_{n\to\infty} a_n$ exists. 

My approach: For any $n\in\mathbb{N},$ $$a_{n}-a_{n+1}=-\frac{1}{n+1}-\ln n+\ln(n+1)\\ =\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}.$$ 
Now we have $$\frac{1}{n+1}<\ln\left(1+\frac{1}{n}\right)<\frac{1}{n}, \forall n\in\mathbb{N}.$$ 
Therefore, $$\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}>0\\ \implies a_{n}-a_{n+1}>0 \\ \implies a_{n}>a_{n+1}, \forall n\in\mathbb{N}.$$ 
This implies that the sequence $\{a_n\}_{n\ge 1}$ is strictly monotonically decreasing. 
Now let us take any $n\in\mathbb{N}$ and approximate the area under the curve $f(x)=\frac{1}{x}, x>0$ represented by the interval $[1,n+1]$. 
Now using definite integral we can approximate the area to be $$A_n=\int_1^{n+1}\frac{dx}{x}=\ln(n+1).$$
Now let us approximate the area using the rectangle method and call that area to be $S_n$. 
To do the same, let us divide the interval into $n$ equal parts in order to have each interval having length $1$. Thus the $n$ intervals will be $[1,2],[2,3],\cdots, [n,n+1].$ Also, we would need to define a function $g:[1,\infty)\to \mathbb{R}$, such that $$g(x)=\frac{1}{[x]}, \forall x.$$ 
Therefore, $$S_n=f(1).1+f(2).1+f(3).1+\cdots+f(n).1\\ =g(1).1+g(2).1+g(3).1+\cdots+g(n).1 \\=\left(\frac{1}{1}\right)1+\left(\frac{1}{2}\right)1+\cdots+\left(\frac{1}{n}\right)1\\=\sum_{i=1}^n\frac{1}{i}.$$
Now since $\forall x\ge 1,$ $$[x]\le x\\ \implies \frac{1}{[x]}\ge \frac{1}{x}([x]=x,\text{ holds true only at integer points, hence }\frac{1}{[x]}=\frac{1}{x},\text{ holds true only at integer points})\\ \implies S_n>A_n\\ \implies \sum_{i=1}^n\frac{1}{i}>\ln(n+1).$$
Therefore, we have $$\sum_{i=1}^n\frac{1}{i}>\ln(n+1)>\ln n\\ \implies\sum_{i=1}^n\frac{1}{i}>\ln n\\ \implies \sum_{i=1}^n\frac{1}{i}-\ln n>0 \\ \implies a_n>0, \forall n\in\mathbb{N}.$$
Therefore, the sequence $\{a_n\}_{n\ge 1}$ is monotonically decreasing and bounded below, hence it is convergent, that is $\lim_{n\to\infty}a_n$ exists and is finite.  
Is the solution correct and rigorous enough, and does there exist any alternate solution to this problem? 
 A: Here is a pictorial answer.  Some people may like it more than a wordy answer.
Let
$$
A_n = \sum_{k=1}^{n-1}\frac{1}{k} - \log n
$$
Since $A_n$ differs from the term in the question by $\frac{1}{n}$ and $\frac{1}{n} \to 0$, it suffices to show that $A_n$ converges.  
Here we have shaded something with area $A_5$.  

The lower graph is $y=1/x$ between $x=1$ and $x=5$.
The
horizontal line segments are:
$y=1$ from $x=1$ to $x=2$;
$y=1/2$ from $x=2$ to $x=3$;
$y=1/3$ from $x=3$ to $x=4$;
$y=1/4$ from $x=4$ to $x=5$.
The area beow the graph of $y=1/x$ is 
$\int_1^5\frac{dx}{x} = \log 5$.
The area below the horizontal line segments is $\sum_{k=1}^4\frac{1}{k}$.
The shaded area is the difference $A_5$.  
Now translate these pieces to the left until they bump the $y$-axis.

The pieces are disjoint, and are contained in the square $[0,1] \times [0,1]$.  When we go from $A_n$ to $A_{n+1}$ we add one more of these pieces.  So the sequence $A_n$ is increasing, and bounded above by $1$.  Therefore $A_n$ converges.
A: Your proof that $a_n>0, \forall n\in\mathbb{N}$ would be correct, except for this line, which you luckily don't use anyway:

This is false : $\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i}$ exists (no!) and is equal to $\int_1^{n+1}\frac{dx}{x}=\ln(n+1).$ 

What you probably meant is "$\sum_{i=1}^n \frac{1}{i} > \ln n$ for all $n\in\mathbb{N}$", let's prove that :
First, notice that 
$$\ln(n) < \ln (n + 1) = \int_1^{n + 1} \frac{1}{x} dx$$
On the other hand, consider the step function $f : x \mapsto \frac{1}{\lfloor x \rfloor}$ , one has $$\sum_{i=1}^n \frac{1}{i} = \int_1^{n + 1} f(x) dx$$
The rest is left to the reader :)
A: You wrote:  $\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i}\text{ exists and is equal to }\int_1^{n+1}\frac{dx}{x}=\ln(n+1).$ But that sum is the famously divergent harmonic series. And also because of that, $S_n$ will not be larger than $\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i}$. 
By comparing the areas (or integral test), $a_n\gt \log(n+1)-\log(n)$ is bounded below. And you have proved $\{a_n\}$ is monotonically decreasing, so that completes the proof.
A: To answer OP's question of whether there is an alternate solution, the most expedite one is to Taylor expand
\begin{align*}
a_{n+1} - a_n & = - \ln \left ( 1 + \frac1n \right ) + \frac{1}{n+1} \\
& = - \frac1n + O \left ( \frac{1}{n^2} \right ) + \frac{1}{n+1} \\
& = - \frac{1}{n(n+1)} + O \left ( \frac{1}{n^2} \right ) \\
& = O \left ( \frac{1}{n^2} \right ).
\end{align*}
Therefore, the series $\sum(a_{n+1}-a_n)$ converges absolutely, and since it is a telescopic series, it means that $(a_n)$ converges.
