Why is this telescoping sum divergent? I'm trying to figure out whether 
$$\sum_{x = 1}^{\infty}\ln\bigg(\frac{x}{x+1}\bigg)$$
converges or diverges.
Here is what I tried doing:
\begin{align}\sum_{x = 1}^{\infty}\ln\bigg(\frac{x}{x+1}\bigg)
 &= \sum_{x = 1}^{\infty}\ln(x)-\ln(x+1) \\
 &= \sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1) \\
 &=(\ln(1) + \ln(2) + \cdots)-(\ln(2)+\ln(3) + \cdots) \\
 &=\ln(1)\end{align}
So I though that the series would converge to $\ln(1) = 0$ since all of the other terms cancel out. I checked using WolframAlpha and the series actually diverges. I was wondering why is this the case and what is wrong with my work above? 
 A: Your attempt is incorrect at this step:
$$
\sum_{x = 1}^{\infty}\ln(x)-\ln(x+1) \color{red}{\overset{?}{=}} \sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1)
$$
In this step, you have rearranged the terms of the series in order to get two new series.  This is not generally possible—see, for example, the Riemann rearrangement theorem.  You make another mistake when you write 
$$
\sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1)
= (\ln(1) + \ln(2) + \cdots)-(\ln(2)+\ln(3) + \cdots)
\color{red}{\overset{?}{=}} \ln(1).
$$
This only makes sense if the two series both converge.  Unfortunately, neither series converges (the series
$ \sum_{n=1}^{\infty} \ln(n)$
diverges to infinity), and your sum-of-series looks like
$$ \sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1) "=" \infty-\infty, $$
which is an indeterminate form.  That is, the expression $\infty-\infty$ is not well-defined, and can take on any value, depending on the underlying construction.
Recall the definition of a convergent series:

We say that
  $$ \sum_{n=1}^{\infty} a_n = S $$
  for some finite $S$ (that is, the series converges to a finite value $S$) if and only if the sequence of partial sums converges to $S$.  That is, with
  $$ S_N := \sum_{n=1}^{N} a_n, $$
  we have $\lim_{N\to\infty} S_N = S$.

We should therefore reason as follows:
\begin{align}
\sum_{x=1}^{\infty} \ln(x)-\ln(x+1)
&= \lim_{N\to\infty} \sum_{x=1}^{N} \ln(x) - \ln(x+1) \\
&= \lim_{N\to\infty} \left[ (\ln(1) - \ln(2)) + (\ln(2)-\ln(3)) + \dotsb (\ln(N) - \ln(N+1)) \right] \\
&= \lim_{N\to\infty} \left[ \ln(1) + (-\ln(2) + \ln(2) + \dotsb (-\ln(N) + \ln(N)) - \ln(N+1) \right] \\
&= \lim_{N\to\infty} \left[ 0 + 0 + \dotsb + 0 - \ln(N+1) \right] \\
&= \lim_{N\to\infty} \left[ -\ln(N+1) \right] \\
&= -\infty.
\end{align}
The important point here is that we are carefully examining the series as a limit of partial sums.  As we do this, we are only using the associativity of addition to get the result, then taking a limit after we have determined the general form of the $N$-th partial sum.  When we do this, we don't lose the "extra" $\log(N+1)$ at the end of the series, which is lost in your formulation.
A: Note that 
$$\sum_{x=1}^X \log(x)-\log(x+1)=\log(1)-\log(X+1)$$
The logarithmic divergence should not be surprising inasmuch as 
$$\log\left(\frac{x}{x+1}\right)=-\log\left(1+\frac1x\right)\le -\frac{1}{x+1}$$
and the harmonic series diverges logarithmically.
