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The ratio test says that if we have

$$\sum_{n=1}^{\infty}a_n$$

such that $\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L$, then if:

1) $L < 1$, then $\sum_{n=1}^{\infty}a_n$ is absolutely convergent,

2) $L > 1$, then $\sum_{n=1}^{\infty}a_n$ is divergent, and

3) $L = 1$, then the ratio test gives no information.

I want to understand the mathematics behind the ratio test. I want to know the reasons behind why and how this works, rather than just memorising a formula.

I have attempted to reason about this theorem myself. It can be seen that we're taking the ratio between the second term in the series, $a_{n+1}$, and the first term in the series, $a_n$. This is exactly how one finds the common ratio in a geometric sequence ($r = \dfrac{a_{n+1}}{a_n} )$.

We then take the limit of this ratio, which I assume is to find the relative rate of change between $a_{n+1}$ and $a_n$. In which case, It is more effective to take the absolute value of the limit, $\lim_{n \to \infty} \begin{vmatrix}{ \dfrac{a_{n+1}}{a_n} }\end{vmatrix} = L$. This seems analagous to how the limit comparison test theorem works. Therefore, I presume that if $L < 1$, this implies that $a_n$ is has a greater rate of change than $a_{n+1}$, which implies that each successive term in the series is getting smaller, and as we go to infinity, each successive term is converging towards $0$. As such, the series should converge to some value. Analogously, I presume that if $L > 1$, this implies that $a_{n+1}$ has a greater rate of change than $a_n$, which implies that each successive term in the series is getting larger, and as we go to infinity, the terms diverge towards infinity.

Is this reasoning correct? If anything is incorrect, please clarify why and what the correct reasoning is.

Thank you.

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    $\begingroup$ "each succesive term is converging towards zero. As such, the series should converge." This is a famously fallacious statement. The harmonic series tends termwise to zero, but the series diverges. $\endgroup$ – Matthew Leingang Nov 11 '16 at 1:13
  • $\begingroup$ @MatthewLeingang I forgot about that. You're absolutely correct. In which case, what is the correct way to reason about that section? $\endgroup$ – The Pointer Nov 11 '16 at 1:14
  • $\begingroup$ @GCab Which is not less than $1$ . . . $\endgroup$ – Noah Schweber Nov 11 '16 at 1:19
  • $\begingroup$ @GCab That's correct, the ratio test doesn't apply to the harmonic series. But the statement I quoted is still invalid. $\endgroup$ – Matthew Leingang Nov 11 '16 at 1:20
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    $\begingroup$ next time read wikipedia's entry en.wikipedia.org/wiki/Ratio_test#Proof $\endgroup$ – reuns Nov 11 '16 at 1:35
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You have some ideas correct, but remember that $a_n \rightarrow 0$ does not imply that $\sum a_n$ converges. So we need something more than that. However, for the case when $L>1$, your reasoning is correct.

For the case $L<1$, we know that there exists some $r < 1$ and $N \in \mathbb{N}$ such that $$\frac{a_{n+1}}{a_n} \leq r < 1,$$ whenever $n \geq N$.

Therefore,

$$ a_{N+1} \leq ra_N, $$ $$ a_{N+2} \leq ra_{N+1} \leq r^2a_N $$ $$ a_{N+3} \leq ra_{N+2} \leq r^3a_N, $$ and in general $$ a_{N+n} \leq r^na_N. $$

Therefore, at least eventually, we can compare the series from above with a convergent geometric series $\sum a_N r^n$, which implies the convergence of the original series $\sum a_n$.

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  • $\begingroup$ I'm not sure if your notation has the same meaning as mine or if you've defined it differently. For instance, how can $\dfrac{ a_{n+1} }{a_n} < r$? I've defined $r = \dfrac{ a_{n+1} }{a_n}$. $\endgroup$ – The Pointer Nov 11 '16 at 1:54
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    $\begingroup$ My notation is different. If you set $r$ equal to the ratio, then it is changing with $n$. But when $L < 1$, such a fixed $r$ exists as I employed it in my answer to your question. $\endgroup$ – user333870 Nov 11 '16 at 2:00
  • $\begingroup$ Can you please elaborate on what $r$ and $N$ are? $\endgroup$ – The Pointer Nov 11 '16 at 2:03
  • $\begingroup$ Since $lim \frac{a_{n+1}}{a_n} = L < 1$, take $\epsilon = \frac{1-L}{2}$, then by the definition of the limit, there exists $N\in \mathbb{N}$ such that $n \geq N \Longrightarrow -\epsilon < \frac{a_{n+1}}{a_n} - L < \epsilon$, so you can take for example $r = \epsilon + L = \frac{1+L}{2}$ $\endgroup$ – user333870 Nov 11 '16 at 2:18
  • $\begingroup$ I'm sorry but this still doesn't tell me what $N$ and $r$ are supposed to be. I cannot see the relevance between your answer and my original question. $\endgroup$ – The Pointer Nov 11 '16 at 5:28

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