Series: Ratio Test

K

KAPPY

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Feb 26, 2013
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4
What am I doing wrong?
We have to determine whether the series converges or diverges and my book says this should converge by the ration test, but I keep getting that the ratio test fails.

summation from n=1 to infinity of [(n+1)*(n+2)]/n!

I plug in n+1 for each n and get: [(n+2)*(n+3)]/[(n+1)*n!]
and I divide by the original equation and I cancel out the (n+2) and n! so I am left with (n+3)/(n+1)^2 which fails the ratio test. What am I doing wrong?
 
KAPPY said:
What am I doing wrong?
We have to determine whether the series converges or diverges and my book says this should converge by the ration test, but I keep getting that the ratio test fails.

summation from n=1 to infinity of [(n+1)*(n+2)]/n!

I plug in n+1 for each n and get: [(n+2)*(n+3)]/[(n+1)*n!]
and I divide by the original equation and I cancel out the (n+2) and n! so I am left with (n+3)/(n+1)^2 which fails the ratio test. What am I doing wrong?
Click to expand...

How did you decide (n+3)/(n+1)^2 fails the ratio test!
 
Subhotosh Khan said:
How did you decide (n+3)/(n+1)^2 fails the ratio test!
Click to expand...

It would be bottom heavy, n/n^2, which when one takes the limit as n approaches infinity would equal 0 so... Oh for some reason I was under the impression that the condition was that if the limit was equal to zero (not one) it fails. I'm sorry for wasting your time.
 
KAPPY said:
It would be bottom heavy, n/n^2, which when one takes the limit as n approaches infinity would equal 0 so... Oh for some reason I was under the impression that the condition was that if the limit was equal to zero (not one) it fails. I'm sorry for wasting your time.
Click to expand...

Absolutely no need to applogize... I made the same mistake several times (that is why it was easy for me guess your trouble)!!!
 
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