Find the convergence region of a series

[Violette]

Find the convergence region of the given function series

[math]\sum_{n=1}^{\infty} \frac{n \cdot 2^n \cdot (x-7)^n}{(n+2)(n+3)}[/math]

 

[mario99]

Did you try the ratio test?

 

[blamocur]

[math]\sum_{n=1}^{\infty} \frac{n \cdot 2^n \cdot (x-7)^n}{(n+2)(n+3)}[/math]

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[Steven G]

Find the convergence region of the given function series

[math]\sum_{n=1}^{\infty} \frac{n \cdot 2^n \cdot (x-7)^n}{(n+2)(n+3)}[/math]

Did you try to simplify anything or do you just want the complete solution posted for you?

 

[Violette]

Did you try the ratio test?

I did try the ratio test and got result of R=1/2, convergence range is between (6.5;7.5), then I tried 7.5 leads to the result of limit = 0 so 7.5 is qualified to be in range, but the problem is I can't solve the x=6.5. Because when I use:
[math]a_n = \frac{(-1/2)^n \cdot n \cdot 2^n}{(n+2)(n+3)} a_n = \frac{(-1)^n \cdot n}{(n+2)(n+3)}[/math][math]b_n = \frac{n}{(n+2)(n+3)}[/math]The problem is when I use the Lebnitz test for alternating series, I found that n go from 1 to infinity but the function is monotomic decreasing only after [math]\sqrt6[/math]. That's why I am kind of concerned about this.

 

[Violette]

Did you try to simplify anything or do you just want the complete solution posted for you?

I showed my work below. I apologize for my poor editing skills

 

[mario99]

I did try the ratio test and got result of R=1/2, convergence range is between (6.5;7.5), then I tried 7.5 leads to the result of limit = 0 so 7.5 is qualified to be in range, but the problem is I can't solve the x=6.5. Because when I use:
[math]a_n = \frac{(-1/2)^n \cdot n \cdot 2^n}{(n+2)(n+3)} a_n = \frac{(-1)^n \cdot n}{(n+2)(n+3)}[/math][math]b_n = \frac{n}{(n+2)(n+3)}[/math]The problem is when I use the Lebnitz test for alternating series, I found that n go from 1 to infinity but the function is monotomic decreasing only after [math]\sqrt6[/math]. That's why I am kind of concerned about this.

I don't think that [imath]7.5[/imath] will be in range. Recalculate.

Using the end points, you will get:

[imath]\displaystyle \sum_{n=1}^{\infty} \frac{n}{(n+2)(n+3)}[/imath]

And

[imath]\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n n}{(n+2)(n+3)}[/imath]

You can solve the first one by limit comparison test and the second by alternating series test.

And don't forget that the idea of using the alternating series test is to show that [imath]a_{n+1} < a_n[/imath], but you did not show that! And the test doesn't have to be true for [imath]n = 1[/imath] or any begining indices. For example, if [imath]a_{n+1} < a_n[/imath] for [imath]n \geq 100[/imath], the series converges.

 

[Violette]

oops my mistake, at x = 7.5 the series will be in the range of divergence, but at 6.5 i have no idea because after doing some research, I saw them ignore n = 1 2 and continue to calculate convergence with n>=3

 

[Violette]

I don't think that [imath]7.5[/imath] will be in range. Recalculate.

Using the end points, you will get:

[imath]\displaystyle \sum_{n=1}^{\infty} \frac{n}{(n+2)(n+3)}[/imath]

And

[imath]\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n n}{(n+2)(n+3)}[/imath]

You can solve the first one by limit comparison test and the second by alternating series test.

And don't forget that the idea of using the alternating series test is to show that [imath]a_{n+1} < a_n[/imath], but you did not show that! And the test doesn't have to be true for [imath]n = 1[/imath] or any begining indices. For example, if [imath]a_{n+1} < a_n[/imath] for [imath]n \geq 100[/imath], the series converges.

Can you help me understand why it doesn't have to be true at n = 1,2

 

[mario99]

oops my mistake, at x = 7.5 the series will be in the range of divergence, but at 6.5 i have no idea because after doing some research, I saw them ignore n = 1 2 and continue to calculate convergence with n>=3

Now this is correct.

For [imath]x = 6.5[/imath], do this:

[imath]\displaystyle \frac{n + 1}{(n + 1 + 2)(n + 1 + 3)} \leq \frac{n}{(n + 2)(n + 3)}[/imath]

Is this inequality true?

 

[mario99]

Can you help me understand why it doesn't have to be true at n = 1,2

Because we don't care about what is happening in the beginning. We only care about what happens at [imath]\infty[/imath].

 

[Dr.Peterson]

Can you help me understand why it doesn't have to be true at n = 1,2

Your textbook should explain this. Here is an example:

Calculus II - Alternating Series Test

In this section we will discuss using the Alternating Series Test to determine if an infinite series converges or diverges. The Alternating Series Test can be used only if the terms of the series alternate in sign. A proof of the Alternating Series Test is also given.

tutorial.math.lamar.edu

It initially says that [imath]b_n[/imath] must be a decreasing sequence, but later says

Secondly, in the second condition all that we need to require is that the series terms, [imath]b_n[/imath] will be eventually decreasing. It is possible for the first few terms of a series to increase and still have the test be valid. All that is required is that eventually we will have [imath]b_n\ge b_{n+1}[/imath] for all n after some point.​

Read the page for details and examples.

 

[blamocur]

Can you help me understand why it doesn't have to be true at n = 1,2

To expand on @mario99's answer: one can ignore any finite number of items in the series because their sum is finite and thus does not effect the convergence.