Tuesday, 5 January 2016

calculus - Does the sequence fracn!2n converge or diverge?



Does the following sequence {an} converge or diverge?




an=n!2n



Answer




Consider writing "out" the sequence:



n!2n=n2n12n2242322212



Note that every time we take another step in the sequence, we multiply by n2 so we're making the sequence larger and larger each time. In particular, we can see that every term in the factorization in (1) is larger or equal than 1, except 1/2, so that



n!2n=n2n12n2242322212



What does this tell you about the limit of the sequence?




Another approach would be D'Alambert's criterion (ratio test), which gives:



{a_n} = \frac{{n!}}{{{2^n}}}



So



\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)!}}{{{2^{n + 1}}}}\frac{{{2^n}}}{{n!}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{2}\frac{{n!}}{{n!}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{2}



What can you say about that limit? Then, what does this tell you about




\mathop {\lim }\limits_{n \to \infty } {a_n} \; \;?


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