Proving Minus euler constant is finite Question
prove that $\lim _{n\to \infty} \left(\ln(n)-\sum_{j=1}^n \frac 1j\right)$ exists and is finite.
Thoughts
We bounded the expression from above with $\ln \frac {n+1}n+\sum\frac 1{2j^2}$. We need to show that the sequence is an ascending monotone sequence. Is this the right way to do it? We don't really know where to start and how (induction?)
 A: Let
$$w_n=\log(n)-\sum_{k=1}^n\frac{1}{k}$$
then the series with general term
$$w_{n}-w_{n-1}=-\log\left(1-\frac{1}{n}\right)-\frac{1}{n}\sim_\infty\frac{1}{2n^2}$$
is convergent so we conclude that the sequence $(w_n)$ is convergent by telescoping.
A: I like a geometrical approach. $\log n$ is the area under the graph of $y=1/x$ from $x=1$ to $x=n$. On the same graph, erect a rectangle of hieght 1 over the interval $[1,2]$; height $1/2$ over $[2,3]$; and so on, to height $1/(n-1)$ over $[n-1,n]$. The area of the rectangles is $\sum_1^{n-1}(1/m)$. The rectangles exceed the area under $y=1/x$ by a bunch of "curvilinear triangles". Slide those triangles to the left, so they all sit over $[1,2]$. They don't fill the area 1 rectangle there (but they do more than half-fill it), so their area, as $n\to\infty$, approaches something between $1/2$ and $1$. 
A: $\int_1^t 1/s ds=\log(t)$  Thus by the same reasoning as in the comparison test, $\sum_{j=1}^n 1/(j+1)\leq\log(n+1) \leq \sum_{j=1}^n 1/j$.  But the difference between the upper and lower bound is $\sum_{j=1}^n 1/(j(j+1))$.  Thus $\log(n+1) -\sum_{j=1}^n 1/j$ is already a bounded sequence.  But it's also monotone decreasing, as one can also verify by the reasoning of writing the log as the integral.  Thus the limit exists with $\log(n+1)$ in place of $\log(n)$.  But that difference also converges to 0, and a sum of convergent sequences converges to the sum.
