# Root test proof using Law of Algebra

• member 731016
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,

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!

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.

member 731016
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 $c < 1$, and set $\epsilon = (1 - c)/2 > 0$. Then by convergence of $|x_n|^{1/n}$ to $c$ there exists $N \in \mathbb{N}$ such that for all $n > N$ we have $$|x_n|^{1/n} < c + \epsilon = \frac{c + 1}{2} \equiv a < 1.$$ (This explains exactly the rrelationship between $N$ and $a$, and why they both exist.)

It follows that for $n > N$ 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 $a_k < b_k$ for $k = 1, \dots, K$ then $$\sum_{k=1}^K a_k < \sum_{k=1}^K b_k.$$ Summing the geometric series using $$a^{K} - 1 = (a - 1)(a^{K-1} + \dots + a + 1)$$ 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 $c > 1$.

member 731016
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 $c < 1$, and set $\epsilon = (1 - c)/2 > 0$. Then by convergence of $|x_n|^{1/n}$ to $c$ there exists $N \in \mathbb{N}$ such that for all $n > N$ we have $$|x_n|^{1/n} < c + \epsilon = \frac{c + 1}{2} \equiv a < 1.$$ (This explains exactly the rrelationship between $N$ and $a$, and why they both exist.)

It follows that for $n > N$ 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 $a_k < b_k$ for $k = 1, \dots, K$ then $$\sum_{k=1}^K a_k < \sum_{k=1}^K b_k.$$ Summing the geometric series using $$a^{K} - 1 = (a - 1)(a^{K-1} + \dots + a + 1)$$ 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 $c > 1$.
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##

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!

fresh_42
They want a lower bound on $|x_n|$, in order to prove that the series diverges. The upper bound is irrelevant.

member 731016
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##.

Last edited:
member 731016

### Similar threads

• Calculus and Beyond Homework Help
Replies
3
Views
225
• Calculus and Beyond Homework Help
Replies
2
Views
893
• Calculus and Beyond Homework Help
Replies
34
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
433
• Calculus and Beyond Homework Help
Replies
1
Views
686
• Calculus and Beyond Homework Help
Replies
7
Views
391
• Calculus and Beyond Homework Help
Replies
13
Views
870
• Calculus and Beyond Homework Help
Replies
13
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
499
• Calculus and Beyond Homework Help
Replies
7
Views
1K