# 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?

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.

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.

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} and by induction

$$a_n where the RHS gives a converging series (it converges to $$\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 $$1$$.

• In the first case, why cannot we choose $r$ itself instead of $s$?. That is if $r<1$, $\exists$ $n_0 \in N$ such that $\frac{a_{n+1}}{a_n}<r<1$ and this implies $a_n <a_0r^n$ and by comparison test the series converges. Commented Jun 27, 2021 at 1:48
• @EkaveeraGouribhatla: in general, your assumption that $\exists n_0\cdots \dfrac{a_{n+1}}{a_n}<r$ is wrong.
– user65203
Commented Jun 28, 2021 at 6:44