Consider the sum
$A = \frac{1}{x-1} + \frac{1}{x-2} + \ldots + 1 = \sum_{n=1}^{x-1}\frac{1}{x-n},\quad x > 2$
Can anyone provide some hints on how to proof that the $\lim_{x\rightarrow\infty}A$ exists or not? Initially I thought the sum goes to infinity as $x$ increases, but plotting $\frac{\partial{A}}{\partial{x}}$ shows that the rate of change of $A$ goes to zero as $x$ increases.
Reindex as follows. $$\sum_{n=1}^{x-1}\frac{1}{x-n} = \sum_{k=1}^{n-1} {1\over k}.$$ This is the harmonic series and it diverges. In fact $$\sum_{k=1}^n {1\over k} = \log(n) + O(1). $$
$A$ is the harmonic series, which diverges due to the fact that we can rearrange it as $$1+\dfrac{1}{2}+(\dfrac{1}{3}+\dfrac{1}{4})+(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8})+\dots\gt 1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dots$$
