Root test proof using Law of Algebra

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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1716961243221.png

My solution is
If ##c < 1##, then let a be a number such that ##c < a < 1 \implies c < a##. Thus for some natural number such that ##n \geq N##

##|x_n|^{\frac{1}{n}} < a## is the same as ## |x_n| < a^n##

By Law of Algebra, one can take the summation of both sides to get,

##\sum_{n = N}^{\infty} |x_n| < \sum_{n = N}^{\infty} a^n## where ##\sum_{n = N}^{\infty} a^n## is by definition a geometric series with ##0 < a < 1## and therefore is convergent.

Since the geometric series is convergent, then by the comparison test, ##\sum_{n = 1}^{\infty} x_n## is absolutely convergent

Now for we consider ##c = 1## case,

Since ##|x_n|^{\frac{1}{n}} > 1## is the same as ##|x_n| > 1^n = 1##

But, ##|x_n| > 1## for all ##n \geq N## which is means that ##\lim_{n \to \infty} |x_n| \neq 0## or that ##\lim_{n \to \infty} x_n \neq 0##

Therefore, by divergence test, ##\sum_{n = 1}^{\infty} x_n## is divergent

Does someone please know whether this is please correct?

I have a doubt whether we are allowed to use the law of algebra for taking the summation of both sides of the inequality, however, I think we can do that since summation it is just a operation (function that takes input, in this case a formula for the nth term, and produces output, in this case alot of numbers) like addition, subtraction etc.

Thanks!
 
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  • #2
For finitely many terms, the ordering remains strict, but for series, the ordering is nonstrict in general. Otherwise, yes, comparison with geometric series is sufficient here.
 
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  • #3
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 346155
My solution is
If ##c < 1##, then let a be a number such that ##c < a < 1 \implies c < a##. Thus for some natural number such that ##n \geq N##

##|x_n|^{\frac{1}{n}} < a## is the same as ## |x_n| < a^n##

You can, and should, make this much more precise.

Suppose first that [itex]c < 1[/itex], and set [itex]\epsilon = (1 - c)/2 > 0[/itex]. Then by convergence of [itex]|x_n|^{1/n}[/itex] to [itex]c[/itex] there exists [itex]N \in \mathbb{N}[/itex] such that for all [itex]n > N[/itex] we have [tex]|x_n|^{1/n} < c + \epsilon = \frac{c + 1}{2} \equiv a < 1.[/tex] (This explains exactly the rrelationship between [itex]N[/itex] and [itex]a[/itex], and why they both exist.)

It follows that for [itex]n > N[/itex] each term of [itex]\sum_{x=N+1}^\infty |x_n|[/itex] is less than the corresponding term of a convergent geometric series, so [itex]\sum_{n=1}^\infty x_n[/itex] converges absolutely.

(It is true, and at this level trivial, that if [itex]a_k < b_k[/itex] for [itex]k = 1, \dots, K[/itex] then [tex]
\sum_{k=1}^K a_k < \sum_{k=1}^K b_k.[/tex] Summing the geometric series using [tex]
a^{K} - 1 = (a - 1)(a^{K-1} + \dots + a + 1)[/tex] before taking the limit [itex]K \to \infty[/itex] avoids any possible difficulty.)

Now for we consider ##c = 1## case,

The case [itex]c = 1[/itex] is inconclusive: consider [itex]\sum_n n^2[/itex] and [itex]\sum_n \frac1{n^2}[/itex].

You have not dealt with the case [itex]c > 1[/itex].
 
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nuuskur said:
For finitely many terms, the ordering remains strict, but for series, the ordering is nonstrict in general. Otherwise, yes, comparison with geometric series is sufficient here.
pasmith said:
You can, and should, make this much more precise.

Suppose first that [itex]c < 1[/itex], and set [itex]\epsilon = (1 - c)/2 > 0[/itex]. Then by convergence of [itex]|x_n|^{1/n}[/itex] to [itex]c[/itex] there exists [itex]N \in \mathbb{N}[/itex] such that for all [itex]n > N[/itex] we have [tex]|x_n|^{1/n} < c + \epsilon = \frac{c + 1}{2} \equiv a < 1.[/tex] (This explains exactly the rrelationship between [itex]N[/itex] and [itex]a[/itex], and why they both exist.)

It follows that for [itex]n > N[/itex] each term of [itex]\sum_{x=N+1}^\infty |x_n|[/itex] is less than the corresponding term of a convergent geometric series, so [itex]\sum_{n=1}^\infty x_n[/itex] converges absolutely.

(It is true, and at this level trivial, that if [itex]a_k < b_k[/itex] for [itex]k = 1, \dots, K[/itex] then [tex]
\sum_{k=1}^K a_k < \sum_{k=1}^K b_k.[/tex] Summing the geometric series using [tex]
a^{K} - 1 = (a - 1)(a^{K-1} + \dots + a + 1)[/tex] before taking the limit [itex]K \to \infty[/itex] avoids any possible difficulty.)



The case [itex]c = 1[/itex] is inconclusive: consider [itex]\sum_n n^2[/itex] and [itex]\sum_n \frac1{n^2}[/itex].

You have not dealt with the case [itex]c > 1[/itex].
fresh_42 said:
Here is an essay about series:
https://www.physicsforums.com/insights/series-in-mathematics-from-zeno-to-quantum-theory/

Maybe you can find some interesting aspects.
Thank you for your replies @nuuskur , @pasmith and @fresh_42!

I found a similar solution online for the proof for ##c > 1##
1718651856339.png


However, I'm confused why they did ##\left|\sqrt[n]{\left|x_n\right|}-c\right| = -(\sqrt[n]{\left|x_n\right|}-c)<c-1## where ##\epsilon = c - 1##. Why did they not consider ##\sqrt[n]{\left|x_n\right|}-c<c-1##. I.e does someone please know why they only consider a specific absolute value expansion?

Thanks!
 
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  • #6
They want a lower bound on [itex]|x_n|[/itex], in order to prove that the series diverges. The upper bound is irrelevant.
 
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  • #7
The root test can be conducted for the quantity ##c:=\limsup \sqrt[n]{|x_n|}##. The advantage being, this quantity always exists, whereas the limit itself need not exist. If ##c>1##, then ##c>1+\delta## infinitely often, hence the general term of the series cannot possibly converge to zero.

Also also, if the ratio test yields ##1##, then the root above will also yield ##1##.
 
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