Problem understanding proof of ratio test The ratio test states that for an infinite series
$$\sum_{n=1}^{\infty} a_n $$
if $\lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} <1$, the series is convergent and if $\lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} >1$, the series is divergent. 
The method I learned to prove the part for convergence is to first prove the absolute convergence of $\sum_{n=1}^{\infty} |a_n| $ for the case of $\lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} <1$.   Then one can say $\sum_{n=1}^{\infty} a_n $ is also convergent. 
The method to prove the divergence part of the ratio test is also to first prove divergence of $\sum_{n=1}^{\infty} |a_n| $ for the case    of $\lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} >1$. Then it was said that  $\sum_{n=1}^{\infty} a_n $ is hence divergent. 
But I am aware that there exists conditionally convergent series where even though $\sum_{n=1}^{\infty} |a_n| $ is divergent, $\sum_{n=1}^{\infty} a_n $ is convergent. 
What am I missing to complete the proof of the divergence part of the ratio test?
 A: The argument you are using for divergence is not correct. If $\lim \frac {|a_{n+1}|} {|a_n|} >1$ then $\{a_n\}$ does not tend to $0$ and hence $ \sum a_n$ is not convergent. This is the standard proof used in ratio test.
A: If $ \lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} >1$, then there is $N \in \mathbb N$ such that  $\frac{|a_{n+1}|}{|a_n|} >1$ for $n >N.$
This gives $|a_{N+k}| > |a_N|>0$  for all $k$.
Conclusion: $(a_n)$ does not converge to $0$.
Hence the series is divergent.
A: The ratio test works by comparison with a geometric series.
Let $r$ be the limit of the ratio. For convenience we omit the absolute values. The two proofs are completely symmetric.
If $r<1$, for sufficiently large $n$ by definition of the limit we have $\dfrac{a_{n+1}}{a_n}<s<1$ and by induction
$$a_n<a_0s^n$$ where the RHS gives a converging series (it converges to $\dfrac{a_0}{1-s}$, which is an upper bound for the series).
If $r>1$, for sufficiently large $n$ we have $\dfrac{a_{n+1}}{a_n}>s>1$ and by induction
$$a_n>a_0s^n$$ where the RHS gives a diverging series.

Note that you can't compare to $a_0r^n$ because the ratio can fluctuate around $r$. But by the definition of the limit, we know that the ratio will eventually remain in an interval that excludes $1$ and is bounded by $s$.
This is precisely for this reason that the test is inconclusive when $r=1$: there is no guarantee that the ratio ever remains on the same side of $1$.
