Ratio test giving convergence, classmate got divergence

[Dobie]

Homework Statement

See attached image. (it should say "ratio" not "ration")

Homework Equations

Ratio series test: An+1/An

The Attempt at a Solution

I have worked this problem over and over and continue to get the same solution. Some guy worked it on the board a couple of days ago and got something different, which our teacher said was right.

I cannot reproduce this and I wonder if the student got it wrong and the professor just didn't check it very well. Either that or I'm losing my mind. There is a test tonight and the problem is from a review Monday.

The answer I continue to get is: 1/2n^2+2n = 1/∞ = 0<1 so it converges.

The other student got: 2n+1/n+1 = 2>1 diverges.

I can post a picture of my work if needed. If I really do have the incorrect answer, can someone please post an image of their work so that I can see what's going on?

Thank you

 

[LCKurtz]

Homework Statement

See attached image. (it should say "ratio" not "ration")

Homework Equations

Ratio series test: A(n+1)/An

The Attempt at a Solution

I have worked this problem over and over and continue to get the same solution. Some guy worked it on the board a couple of days ago and got something different, which our teacher said was right.

I cannot reproduce this and I wonder if the student got it wrong and the professor just didn't check it very well. Either that or I'm losing my mind. There is a test tonight and the problem is from a review Monday.

The answer I continue to get is: 1/(2n^2+2n) = 1/∞ = 0<1 so it converges.

The other student got: (2n+1)/(n+1) = 2>1 diverges.

I can post a picture of my work if needed. If I really do have the incorrect answer, can someone please post an image of their work so that I can see what's going on?

Thank you

Your answer is incorrect. And the way you are sloppy with parentheses, it's no surprise. Post your work if you want us to help you.

 

[Dobie]

Well thanks for the advice on using parentheses. I do not believe the formula should have them as you have in your correction, though.

Was the other student's answer correct? Did you solve the problem or just assume I missed it based on my sloppy use of (or lack thereof) parentheses?

My work:

Edit: went back and did not distribute the 2.

 

[LCKurtz]

Well thanks for the advice on using parentheses. I do not believe the formula should have them as you have in your correction, though.

Was the other student's answer correct?

Yes.

It's hard to follow what you did but I think you have calculated$$
(2(n+1))! = (2n+2)n!$$which isn't correct.

 

[Dobie]

Yes.

It's hard to follow what you did but I think you have calculated$$
(2(n+1))! = (2n+2)n!$$which isn't correct.

Even if I do not distribute the 2 (I pulled the 2 back out toward the end), I am still left with 1/∞. Thanks for illustrating the correct way to solve the problem, though!

 

[xiavatar]

$(2(n+1))!=(2n+2)(2n+1)((2n)!)$

 

[Dobie]

$(2(n+1))!=(2n+2)(2n+1)((2n)!)$

Thank you so much for this explanation. I did not know it expanded like that in this case, but I've reworked it and it comes out perfectly.

 

[xiavatar]

Just note that for next time that (a(n+1))!=(an+a)!=(an+a)(an+a-1)(an+a-2)...(2)(1). Always remember to carry out the operation in the parentheses before you do the operations outside of the parentheses. I'm guessing that's why you factored out a 2. You cannot fact out the 2 like that when your doing a factorial.

Examples.
(2*3)!=(6)!=6*5*4*3*2*1.
(4*3)!=(12)!=12*11*10*...*3*2*1