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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?

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3 Answers 3

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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.

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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.

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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 $1$.

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  • $\begingroup$ 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. $\endgroup$ Commented Jun 27, 2021 at 1:48
  • $\begingroup$ @EkaveeraGouribhatla: in general, your assumption that $\exists n_0\cdots \dfrac{a_{n+1}}{a_n}<r$ is wrong. $\endgroup$
    – user65203
    Commented Jun 28, 2021 at 6:44

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