Math Is Fun Forum

EMPhillips1989

Member

Registered: 2008-01-21

Posts: 40

series

does anyone know if this series converges or diverges?
What test for convergence would be best to use for this type of series??

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: series

First you should simplify the expression.  Notice that you can remove one n from the expression and get n/(n-1)! as your series.  I would then separate the second n from the expression by making another faction that is less than 1:

The second fraction will be less than 1 for all n > 3, so it can be dropped for the purpose of determining convergence.  We're now left with the more simple summation of 1/(n-3)!.  Since we're testing for convergence all that matters is how the sum behaves as n approaches infinity, so in order to simplify the expression further we can remove the -3 portion in the denominator to leave us with the following expression:

From here do a comparison test.  Find a series that you know converges and show that it is greater than your series.  This will prove that your series converges.  My advice would be to go with 1/2^n.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: series

Comparison test is not really needed as your last summation is equal to e.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: series

Can't you just do a ratio test from the start?

             

The ratio between terms of the sequence you're summing converges to less than 1, and so their summation converges also.

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Why did the vector cross the road?
It wanted to be normal.