How do you apply the ratio test to determine if #Sigma (3^n(n!))/(n^n)# from #n=[1,oo)# is convergent to divergent?

Answer 1

It is divergent.

We can apply d'Alembert's ratio test:

Suppose that;

# S=sum_(r=1)^oo a_n \ \ #, and #\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| #

Then

if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

# S=sum_(n=1)^oo (3^n(n!))/n^n #

So our test limit is:

# L = lim_(n rarr oo) | {(3^(n+1)((n+1)!))/(n+1)^(n+1)}/{(3^n(n!))/n^n}| # # \ \ \ = lim_(n rarr oo) | (3^(n+1)(n+1)!)/(n+1)^(n+1) * n^n/(3^n n!}| # # \ \ \ = lim_(n rarr oo) | (3 * 3^n(n+1) * n!)/(n+1)^(n+1) * n^n/(3^n n!}| # # \ \ \ = lim_(n rarr oo) | (3 (n+1))/(n+1)^(n+1) * n^n| # # \ \ \ = lim_(n rarr oo) | (3n^n )/(n+1)^n | # # \ \ \ = 3 \ lim_(n rarr oo) | (n/(n+1))^n | # # \ \ \ = 3 \ lim_(n rarr oo) (n/(n+1))^n # # \ \ \ = 3/e # # \ \ \ ~~ 1.1 > 1 #

Hence the series is divergent.

In case you are wondering how the final limit was established; it is a standard limit in disguise:

Let #u=n+1 => n=u-1# then #u rarr oo# as #n rarr oo#, And
# lim_(n rarr oo) (n/(n+1))^n = lim_(u rarr oo) ((u-1)/u)^(u-1)# # " " = lim_(u rarr oo) (1-1/u)^u / (1-1/u)# # " " = {lim_(u rarr oo) (1-1/u)^u }/ {lim_(u rarr oo)(1-1/u)}# # " " = lim_(u rarr oo) (1-1/u)^u #
Which courtesy of Leonhard Euler is standard limit with known value #1/e#.
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Answer 2

See below.

Using Strirling approximation

#log n! =nlogn-n+O(logn)#

we have

#(3^n(n!))/(n^n) approx (3/e)^n# but as #3/e > 1# the series diverges.
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Answer 3

To apply the ratio test to the series ( \sum_{n=1}^{\infty} \frac{3^n(n!)}{n^n} ), we compute the limit of the ratio of consecutive terms:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{3^{n+1}((n+1)!)}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n(n!)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{3^{n+1}(n+1)(n!)}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n(n!)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{3(n+1)}{(n+1)^n} \right| ]

[ = \lim_{n \to \infty} \left| \frac{3}{(1+\frac{1}{n})^n} \right| ]

[ = \lim_{n \to \infty} \left| 3 \right| = 3 ]

Since the limit is greater than 1, the series is divergent by the ratio test.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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