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While going through Analysis 1 textbook by V. A. Zorich, I encountered this proof of d’Alembert’s test for convergence, which has this one part whose purpose isn't very clear to me.

The statement and the proof:

Suppose the limit $\lim_{n \to \infty} \left\lvert \frac {a_{n+1}}{a_n}\right\rvert=\alpha$ exists for the series $\sum_{n=1}^{\infty}a_n$. Then,

a) if $\alpha < 1$, the series $\sum_{n=1}^{\infty}a_n$ converges absolutely;

b) if $\alpha > 1$, the series $\sum_{n=1}^{\infty}a_n$ diverges

c) there exist both absolutely convergent and divergent series for which $\alpha=1$.

Proof.

a) If $\alpha<1$, there exists a number $q$ such that for $\alpha<q <1$. Fixing $q$ and using properties of limits, we find an index $N \in \mathbb{N}$ such that $\left\lvert \frac {a_{n+1}}{a_n}\right\rvert<q$ for $n>N$. Since a finite number of terms has no effect on the convergence of a series, we shall assume without loss of generality that $\left\lvert \frac {a_{n+1}}{a_n}\right\rvert<q$ for all $n ∈ \mathbb{N}$.

Since

$$\underbrace{\left\lvert \frac {a_{n+1}}{a_n}\right\rvert \cdot \left\lvert \frac {a_n}{a_{n-1}}\right\rvert \dots \left\lvert \frac {a_2}{a_1}\right\rvert=\left\lvert \frac {a_{n+1}}{a_1}\right\rvert}_{\text{The problematic part}} $$

we find that $|a_{n+1}| ≤ |a_1| · q^n$. But the series$\sum_{n=1}^{\infty}|a_1| q^n$ converges (its sum is obviously $\frac{|a1|q}{1−q}$), so that the series $\sum_{n=1}^{\infty}a_n$ converges absolutely.

Since I have no issues with parts for $\alpha > 1$ and $\alpha = 1$ I will skip those.

End of proof.


My question:

The issue is, I don't understand what is the role of the underbraced "Problematic part". I am aware of the processes which occur there (a lot of fractions cancel out so we are left only with $\left\lvert \frac {a_{n+1}}{a_1}\right\rvert$) and that we use the comparison test later to finish the reasoning. But why are we looking at the product of all terms of the $\left\lvert \frac {a_{n+1}}{a_n}\right\rvert$ sequence ?

Were the both sides of the inequality $\left\lvert \frac {a_{n+1}}{a_n}\right\rvert<q$ perhaps raised to the $n$-th power and then we somehow got $|a_{n+1}| ≤ |a_1| · q^n$ ? If no, what is happening from this step onward ?

EDIT: Another phrasing of my question: Why did it make sense at that point in the proof, to get the idea to observe the product of successive terms of an+1an and why is the way of arriving from there to |an+1|<r|an| so 'obvious' that no explanation was given before that jump in deducing ?

Thanks

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  • $\begingroup$ Maybe just multiply both sides of the problematic part by $|a_1|$. Then use that $ \frac{|a_{n+1|}}{|a_n|} < q $ for all natural number n. $\endgroup$
    – PAM1499
    Commented Aug 26, 2020 at 19:52

2 Answers 2

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Using all the symbols as in your post and considering only the case $\alpha \lt 1$, try to understand the proof like this (i.e., don't just assume at the very beginning that $\left\lvert \frac {a_{n+1}}{a_n}\right\rvert<q$ is true for all $n\in \mathbb N$):
$\left\lvert \frac {a_{n+1}}{a_n}\right\rvert<q$ for $n\gt N$
Therefore, $|a_{n+1}|\lt q|a_n|\lt q(q|a_{n-1}|)=q^2a_{n-1}\lt\cdots\lt q^{n-N}|a_{n-(n-(N+1))}|=q^{n-N}|a_{N+1}|\implies |a_{n+1}|\lt|a_{N+1}q^{-N}||q^{n}|\tag{1}$
$\sum_{n=N+1}^{\infty}|a_{N+1}q^{-N}||q^{n}|$ converges as $|q|\lt 1$ and hence by comparison test the series $\sum_{n=N+1}^{\infty}|a_{n+1}| $ also converges and adding a finite no. of terms won't affect its convergence and therefore $\sum_{n=0}^{\infty}|a_{n+1}| $ also converges.

Reply to your queries:

"But why are we looking at the product of all terms of the $|\frac{a_{n+1}}{a_n}|$sequence ?" - Because we want to compare $|a_n|$ with a convergent sequence which we get as per $(1)$.

"Were the both sides of the inequality $\left\lvert \frac {a_{n+1}}{a_n}\right\rvert<q$ perhaps raised to the n-th power and then we somehow got $|a_{n+1}| ≤ |a_1| · q^n$ ?"- Please refer $(1)$ how $q$ raised to power $n$ was obtained.

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  • $\begingroup$ How do we know (or why is it promising) that the product of successive terms of $|\frac{a_{n+1}}{a_n}|$ is going to help us in finding the sequence to compare with $|a_n|$? I am basically asking what in the proof is an indicator that observing that product would be a good idea for our purpose. I understand your reasoning in $(1)$, and how it gets us to the final conclusion. What I fail to understand is, for what choices of $N$ and/or $n$ in $|a_{n+1}|\lt|a_{N+1}q^{-N}||q^{n}|$ do we get $|a_{n+1}| ≤ |a_1| · q^n$? If that is what your second reply to my queries was implying $\endgroup$
    – powerline
    Commented Aug 26, 2020 at 23:54
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    $\begingroup$ @GrigoriPerelman: For what choices of $N$? $N$ exists because limit of the ratio exists (limit is equal to $\alpha$) and depends upon the $\epsilon$ (arbitrary difference between $\alpha$ and $|a_{n+1}/a_n| )$ you choose. As for the indicator part, $\alpha \lt 1$ indicates "roughly" that for "sufficiently" large values of $n$, the ratio of successive terms is less than 1 that is, in more words $(n+1)$th term is getting smaller than $a_n$. By this time, it is observed by ratio test that $a_n\to 0$ (necessary condition for $\sum x_n$ to be convergent) so comparison test confirms the rest. $\endgroup$
    – Koro
    Commented Aug 27, 2020 at 0:04
  • $\begingroup$ I took another look at your original answer; how do we know that the condition for geometric series $|q|<1$ is enough for this series $\sum_{n=N+1}^{\infty}|a_{N+1}q^{-N}||q^{n}|$ to converge? I am not that familiar with the matter, to my understanding it isn't neccessary that it converges since we have those additional terms $|a_{N+1}q^{-N}|$. You can just send me a link with the rule if there is any. Many thanks $\endgroup$
    – powerline
    Commented Aug 27, 2020 at 1:17
  • $\begingroup$ Is perhaps $|a_{N+1}q^{-N}|$ a constant value ? If yes then everything is clear. $\endgroup$
    – powerline
    Commented Aug 27, 2020 at 1:27
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    $\begingroup$ @GrigoriPerelman: Indeed $ |a_{N+1}q^{-N}|$ is a constant value. Why? Because $N$ is a constant. Why? Because by limit definition, there exists an $N$ such that for all n$\gt N$, the ratios are bounded by $q$. To understand that, you supply the arguments why lim (ratio) =$\alpha\implies $lim(ratio) $\lt q$ $\endgroup$
    – Koro
    Commented Aug 27, 2020 at 3:41
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Maybe it's easier to see this if we just take it in steps:

$1).\ $ There is an $0<r<1$ and an integer $N$ such that $\left|\frac{a_{n+1}}{a_n}\right|<r$ whenever $n>N.$ Fix this $N$ and $r$. Then,

$2).\ |a_{n+1}|<r|a_n|$ for $n>N.$ So,

$3).\ |a_{N+1}|<r|a_N|$

$\ |a_{N+2}|<r|a_{N+1}|<r^2|a_N|$

$|a_{N+3}|<r|a_{N+2}|<r^3|a_N|.$

Now we see the pattern, so

$4).\ |a_{N+k}|<r^k|a_N|$. This means that

$5).\ \sum_{k=1}^\infty |a_{N+k}|$ converges by the comparison test. But this series is just a tail of the original series of absolute values, $\sum_{n=1}^\infty |a_{n}|$, which means that

$6).\ \sum_{n=1}^\infty a_n$ converges absolutely.

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  • $\begingroup$ So what is the purpose of the product then in the "Problematic part" if $\ |a_{N+k}|<r^k|a_N|$ ? And if $|a_{n+1}<r|a_n| $ for n>N how can we say that $|a_{N+1}<r|a_N|$ since $N<n$, and we don't know if the sequence is increasing or decreasing ? $\endgroup$
    – powerline
    Commented Aug 26, 2020 at 23:14
  • $\begingroup$ The problematic part is just another way of arriving at the conclusion in my step 4. It doesn't matter whether the sequence is decreasing or not. If you review steps 2 and 3, you will see that it's just a matter of successive back-substitution, and the fact that $0<r<1.$ $\endgroup$ Commented Aug 26, 2020 at 23:26
  • $\begingroup$ Well, although I appreciate your insightful and very helpful answer, my question was how did we arrive at $|a_{n+1}| ≤ |a_1| · q^n$ using the underbraced product. Even with your reasoning I still can't see how we got there, although I fully understand it and it makes perfect sense. Except I am not fully aware how to get the initial idea where we are supposed to look how the inequalitiy $\ |a_{n+1}|<r|a_n|$ behaves for $n+1, n+2,...$ and the intuition behind it. $\endgroup$
    – powerline
    Commented Aug 26, 2020 at 23:35
  • $\begingroup$ Another phrasing of my question: Why did it make sense at that point in the proof, to get the idea to observe the product of successive terms of $\frac{a_{n+1}}{a_n}$ and why is the way of arriving from there to $\ |a_{n+1}|<r|a_n|$ so 'obvious' that no explanation was given before that jump in deducing ? $\endgroup$
    – powerline
    Commented Aug 26, 2020 at 23:41
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    $\begingroup$ Because each factor is a quotient of terms $<q$ and since they all reduce (by cancellation) to the right-hand side, you get immediately that $|a_{n+1}| ≤ |a_1| · q^n$. I find this somewhat confusing however, which is why I offered a different approach. $\endgroup$ Commented Aug 26, 2020 at 23:48

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