- #1

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!

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!