Interval of Convergence

[I Love Pi]

I circled two ways I tried in determining the interval of convergence. Could you please let me know which method proves the interval of convergence is at the point 1/2? Thank you in advance!

 

[Steven G]

In calculating p you replaced (n+1) with1 + 1/n. Can you justify that step?

(-2)^n is an alternating series with the terms getting larger and larger. The difference between two consecutive terms is growing without bound. That along with n! growing w/o bound should tell you that n!(-2)^n is not convergent.

Just calculate the p value correctly and be done.

 

[I Love Pi]

My solutions manual says the point of convergence is {1/2}. If p diverges as you suggest, then wouldn’t this conflict with the solutions manual?

 

[Steven G]

For which values of x is p in [0,1). Recall \(\displaystyle p=\lim_{n\rightarrow\infty}(n+1)!|2x-1|\)

Note that in this limit, |2x-1| is simply a constant.

 

[I Love Pi]

This was very helpful. I have attached my revised solution. Please let me know if this is what you meant. Thank you, again!

 

[Steven G]

This was very helpful. I have attached my revised solution. Please let me know if this is what you meant. Thank you, again!

You have multiple answers! You wrote that |2x-1|*oo = oo. This is not true! If x = 1/2, then |2x-1|=0. 0*(something approaching infinity) does not equal infinity, it equal 0! Just like you said at some point.

 

[Steven G]

My solutions manual says the point of convergence is {1/2}. If p diverges as you suggest, then wouldn’t this conflict with the solutions manual?

Yes, but who said that p diverges for all values of x.