How to show $\gamma$ aka Euler's constant is convergent? I'm trying to show different convergents, and this is the first one i'm having problems with. 
$\dfrac11 + \dfrac 12 + \ldots + \dfrac 1n - \log n \rightarrow \gamma$ 
Like it's the definition of euler's constant, but how to show that this expression is convergent( i don't mean convergent to specific number, just convergent ). 
I tried d'alambert's criterion. This is : if $$\left| \frac{a_{n+1}}{a_{n}}\right | < 1$$ then $a_{n}$ is convergent. 
But i end up with something like 
$$\frac{H_{n} + (n+1) - ln(n+1)}{H_{n} - \ln(n)}$$ where $H_{n}$ is n-th harmonic number. I have no clue how to work on it further. 
Would love to get some hints or solutions on this.  
Cheers
 A: $\gamma$ is the limit of the sum of the slightly bigger than triangle pieces of this diagram (from Wikipedia)

As $n$ increases, the sum increases, but clearly has an upper bound of $1$ and therefore converges to a limit $\gamma$ less than or equal to $1$. 
This picture also makes it obvious why $\gamma$ is slightly more than $0.5$
In fact the partial sum of the pieces is $H(n)-\log_e(n+1)$ but the difference $\log_e(n+1) - \log_e(n)$  is $O(\frac1n)$, so does not affect the convergence to $\gamma$.
A: Let $$u_n=\sum_{k=1}^n\frac 1 k-\log n$$
then
$$u_{n}-u_{n-1}=\frac{1}{n}+\log\left(1-\frac 1 n\right)\sim_\infty-\frac{1}{2n^2}$$
so the series $\displaystyle\sum_{n\ge2}u_{n}-u_{n-1}$ is convergent by asymptotic comparison and then the sequence $(u_n)_n$ is convergent by telescoping.
A: the sequence $U_n$is strictly decreasing sequence.
as,    $U_{n+1}-U_n=\frac{1}{n+1}-\log  (1+\frac{1}{n})<0$;which can be checked by Riemann's integral on the function $f(x)=\frac{1}{x}$.
now again using Riemann's integral we can show,
$\frac{1}{n}>\log(n+1)-\log(n)$,so,
$$\sum_{k=1}^{n}\frac{1}{k}>(\log 2-\log 1)+(\log 3-\log 2)+(\log 4-\log 3)+....+(\log(n+1)-\log n )=\log(1+n)$$
$$U_n>\log(1+n)-\log n>0;$$
so,$U_n$ has a lower bound as well as decreasing,hence $U_n$ converges.
