# Determine whether the given series converges or diverges. `sum_(n=0)^oo` `(- 3)^(2n)/(n!)`

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### 1 Answer

`sum_(n=0)^oo` `(-3)^(2n)/(n!)`

Since the given series involves a factorial, we are going to use ratio test to determine if it is divergent or convergent.

In ratio test, the formula is :

`L= lim_(n->oo) a_(n+1)/a_n`

where `a_(n+1)` and `a_n` are the *(n+1)*th and *n*th terms of the series. When L<1, the series is convergent. And if L>1, the series is divergent.

Then, let's determine *(n+1)*th term of the series.

`a_(n+1) = (-3)^(2(n+1))/((n+1)!)=(-3)^(2n+2)/((n+1)!) = ((-3)^(2n)*(-3)^2)/((n+1)n!) = (9*(-3)^(2n))/((n+1)n!)`

And the *n*th term is:

`a_n = (-3)^(2n)/(n!)`

Divide `a_(n+1)` by `a_n` .

`a_(n+1)/a_n = [(9*(-3)^(2n))/((n+1)n!)]/[(-3)^(2n)/(n!)] = (9*(-3)^(2n))/((n+1)n!) * (n!)/(-3)^(2n)`

`a_(n+1)/a_n= 9/(n+1)`

Take the limit of `a_(n+1) / a_n` as *n* approaches infinity.

`L= lim_(n->oo)` `a_(n+1)/a_n` `=` `lim_(n->oo)` `9/(n+1)`

In taking the limits of a rational function at infinity, we need to determine the term with the highest degree (or variable with highest exponent). Then, divide all the terms by the term with the highest degree.

The term with the highest degree in the above rational function is *n*. So, we have,

`= lim_(n->oo) ` `(9/n)/(n/n+1/n)` `=` `lim_(n->oo)` `(9/n)/(1+1/n)`

Then, take the limit of each term separately.

For fractions, apply the rule `lim_(x->oo) 1/x=0.`

>> `lim_(n->oo) 9/n =0`

>> `lim_(n->oo) 1/n = 0`

For the constant, its limit is equal to the constant itself

>> `lim_(n->oo) 1 = 1`

So we have,

`L =` `lim_(n->oo)` `(9/n)/(1+1/n) = 0/(1+0) = 0`

**Since L<1, the series `sum_(n=0)^oo (-3)^(2n)/(n!)` is convergent.**