Student Question

`sum_(n=0)^oo 2^n/(n!)` Use the Root Test to determine the convergence or divergence of the series.

Expert Answers

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It is usually easier to use ratio test on these types of series that contain factorials. However, we can also use root test if we rewrite factorial using exponentials. This can be accomplished using Stirling's approximation

`n! approx sqrt(2pi n)(n/e)^n`

The reason why we can use this approximation is because it becomes more precise for greater values of `n,` in fact the ratio of the left and right hand side of the approximation converges to 1 as `n` tends to infinity.

`lim_(n to infty)root(n)(2^n/n!) =lim_(n to infty) root(n)(2^n/(sqrt(2pi n)(n/e)^n))=lim_(n to infty)2/(root(n)(sqrt(2pi n))n/e)=`

In order to calculate `lim_(n to infty) root(n)(sqrt(2pi n))` we need to use the following two facts:

`lim_(n to infty) root(n)(c)=1,` `c in RR` and `lim_(n to infty)root(n)(n^p)=1,` `p in RR.`

                                                                                                                    Applying this to our limit yields

`lim_(n to infty)2/(root(n)(sqrt(2pi n))n/e)=lim_(n to infty)2/(n/e)=lim_(n to infty)(2e)/n=(2e)/infty=0`

Since the value of the limit is less than 1, the series is convergent. 

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