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

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

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

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

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

You have not dealt with the case c &gt; 1.
 
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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 c &lt; 1, and set \epsilon = (1 - c)/2 &gt; 0. Then by convergence of |x_n|^{1/n} to c there exists N \in \mathbb{N} such that for all n &gt; N we have |x_n|^{1/n} &lt; c + \epsilon = \frac{c + 1}{2} \equiv a &lt; 1. (This explains exactly the rrelationship between N and a, and why they both exist.)

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

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



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

You have not dealt with the case c &gt; 1.
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!
 
They want a lower bound on |x_n|, in order to prove that the series diverges. The upper bound is irrelevant.
 
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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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