Test on the difference of reciprocals I am trying to prove that a certain series $\sum a_n$ diverges.
All I know is that the terms are positive, strictly decreasing and tend to zero. 
Moreover  the two following limits hold:
$$\lim_{n \rightarrow   \infty} \frac{a_{n+1}}{a_n} = 1$$ 
$$\lim_{n \rightarrow   \infty} \frac{1}{a_{n+1}}-\frac{1}{a_n}= \frac{1}{3}$$
Is there any standard test that comes to your mind?
I have been wondering if the following is true: 
if $\sum a_n$ converges then $\frac{1}{a_{n+1}}-\frac{1}{a_n}$ is unbounded
Not sure if this holds but it would solve the problem as well.
Thank you very much for any hint/suggestion you might have! 
 A: Hint:  here is a heuristic argument:
Using a telescoping series, $\frac{1}{a_n} - \frac{1}{a_1} \approx \frac{n}{3}$.  Taking reciprocals, 
$a_n \approx \frac{3}{n}$, so the series behaves like the harmonic series, which diverges.  You must fill in some missing steps, probably by showing your series is greater than some constant multiple of the harmonic series.
To make this more rigorous, one way would be to use the fact that $\frac{1}{a_{n+1}} - \frac{1}{a_n} < \frac{1}{2}$ for $n \geq N$ (for some $N \in \mathbb{N}^+$), use a telescoping series to bound $1/a_n$ from above, and hence $a_{n}$ from below for large $n$, and compare $a_n$ to some multiple of $1/n$ for large $n$, thereby showing that a "tail" of the series diverges.  A comment below suggests using the Limit Comparison Test.  If you can get this to work, it would probably lead to a cleaner and more elegant solution. 
The assumption that the terms in the series are decreasing seems to be unnecessary. 
Your idea ("I have been wondering...") is an interesting one.  Doubtless someone will prove it one way or the other soon.
I don't know if any of the "standard tests" work.  Obviously the Ratio Test fails.
