Use the ratio test to determine whether sumn=4^infty...

Use the ratio test to determine whether sum_{n=4}^{infty} frac{n(-3)^n}{n!} converges or diverges. (a) Find the ratio of successive terms. Write your answer as a fully simplified fraction. For n ge 4, lim_{n o infty} left| frac{a_{n+1}}{a_n} ight| = lim_{n o infty} frac{quad}{quad} (b) Evaluate the limit in the previous part. Enter infty as infinity and -infty as -infinity. If the limit does not exist, enter DNE. lim_{n o infty} left| frac{a_{n+1}}{a_n} ight| = (c) By the ratio test, does the series converge, diverge, or is the test inconclusive? Choose

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00:01 Hi, let's start the solution in this question.

00:03 We have given series summation n is 4 to infinity n minus 3 power n divided by n factorial so here we have a n is n minus 3 power n divided by n factorial and we know that n factorial can be written as n n minus 1 so many terms then a n plus 1 is n plus 1 into minus 3 power n plus 1 divided by n plus 1 factorial now then simplify this term that is a n plus 1 which one is minus 3 power n into minus 3 n plus 1 use definition of n plus 1 factorial that is n plus 1 into n factorial this one is cancelled out so we get here a n plus 1 is minus 3 power n into minus of 3 divided by n factorial then by ratio test limit n tends to infinity a n plus 1 divided by a n which is equal to 0 sorry, which is equal to limit? we find the limit that is limit n tends to infinity a n plus 1 is minus 3 power n minus 3 divided by n factorial and whole divided by a n which is n minus 3 power n divided by n factorial so we get that is equal to limit n tends to infinity mod minus 3 by n which is equal to minus 3 by infinity that is equal to 0 then we get limit n tends to infinity mod a n plus 1 divided by a n which is equal to 0 which is less than 1 then by ratio test given series summation n is 4 to infinity n minus 3 power n divided by n factorial is absolutely convergent absolutely convergent that implies series is convergent hence we get series summation n is 4 to infinity n minus 3 power n divided by n factorial is convergent now answer of part a that is limit n tends to infinity mod a n plus 1 divided by a n which is equal to limit n tends to infinity 3 by n so we get here limit n tends to infinity mod a n plus 1 is equal to limit n tends to infinity minus 3 by n answer of part b that is limit n tends to infinity mod a n plus 1 upon a n that is equal to 0 and c by the ratio test that is the answer of part c does the series converges diverges or the test inconclusive so by previous one step we get limit n tends to infinity mod a n plus 1 divided by a n which is equal to 0 which is less than 1 that implies by ratio test series is converges that is the answer...

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