Finding The Radius Of Convergence.

[chlxowns_]

Introduction

Hello, I am currently a student in Engineering Degree, and currently, I am stuck in this particular question regarding finding the radius of convergence using the ratio test... Any help is appreciated!

Question

Find the Radius of convergence:



My Attempt:




I am currently stuck here, as the answer is given is not the answer I found...

 

[skeeter]

fyi, [MATH]\lim_{n \to \infty} \dfrac{\infty}{\infty} \ne 1[/MATH]
I get [math]\dfrac{1}{7}|x-2| < 1 \implies |x-2| < 7[/math]

 

[HallsofIvy]

One obvious point- since the ratio test requires that the terms be positive, you have to take the absolute value so you should have something like "|x|< 1" and so -1< x< 1, NOT "x< 1".

 

[Cubist]

You wrote...



The first line is good, but the next line is where the problem started because it isn't a good idea to substitute n=∞. The reason for the limit notation is that n can't equal infinity, since infinity isn't a number. (However n can tend towards infinity.)

A better approach is to expand the numerator to (n^3 + 3n^2 + 3n + 1)(5^n + 7^n). Then, divide both the numerator and denominator by n^3*7^n. After this, many of the terms in the numerator and denominator will tend towards 0 as n tends towards infinity. Please post back if you don't understand.

With practice, you'll just spot the dominant terms in the numerator and denominator, and dividing their coefficients will give the result of the limit directly (without having to divide everything).

 

[Steven G]

Recall that the limit of a product is the product of the limit.

I would break up your limit into [MATH]\lim_{n\to\infty}\dfrac{({n+1})^3}{{n^3}}[/MATH]*[MATH]\lim_{n\to\infty}\dfrac{{5^n+7^n}}{5^{n+1} + 7^{n+1}}[/MATH]


Now the first limit is 1. We would like (x-2) times the 2nd limit to be less that or equal to one.

See what you can do with that.

[/QUOTE]

 

[skeeter]

Recall that the limit of a product is the product of the limit.

I would break up your limit into [MATH]\lim_{n\to\infty}\dfrac{({n+1})^3}{{n^3}}[/MATH]*[MATH]\lim_{n\to\infty}\dfrac{{5^n+7^n}}{5^{n+1} + 7^{n+1}}[/MATH]


Now the first limit is 1. We would like (x-2) times the 2nd limit to be less that or equal to one.

less than 1 ... ratio test = 1 is inconclusive

one may check the endpoints to determine if they are within the interval of convergence

 

[chlxowns_]

With practice, you'll just spot the dominant terms in the numerator and denominator

Hi! Thank you for your reply! May I know if there is a way to easily spot the dominant term?

I would break up your limit into limn→∞(n+1)3n3limn→∞(n+1)3n3\displaystyle \lim_{n\to\infty}\dfrac{({n+1})^3}{{n^3}}*limn→∞5n+7n5n+1+7n+1limn→∞5n+7n5n+1+7n+1\displaystyle \lim_{n\to\infty}\dfrac{{5^n+7^n}}{5^{n+1} + 7^{n+1}}

Thank you I will try that!

 

[chlxowns_]

One obvious point- since the ratio test requires that the terms be positive, you have to take the absolute value so you should have something like "|x|< 1" and so -1< x< 1, NOT "x< 1".

Noted! Thank you for the clarification!

 

[chlxowns_]

fyi, limn→∞∞∞≠1limn→∞∞∞≠1\displaystyle \lim_{n \to \infty} \dfrac{\infty}{\infty} \ne 1

Ah Thank you for telling me!

 

[Cubist]

May I know if there is a way to easily spot the dominant term?

I wouldn't recommend learning a set of rules for spotting the dominant term. It's just something that will seem commonsense to you once you've gained some extra experience. Practice, practice. For now there's a big advantage to doing it the slow way - you might pick up method marks in the the case that you accidentally make a mistake!

Certainly use the advice given in post#5 to break some products up (while doing this look for similar types of terms in the numerator and denominator and keep them together). The idea is to divide numerator and denominator by the same amount in order to get most terms tending toward zero.

EDIT: Having said that, in the example below just think about the binomial theorem to quickly find the dominant term in the numerator (without fully expanding)

[MATH]\lim_{n\to\infty}\dfrac{({n+1})^3}{{n^3}}[/MATH]