Show that the sequence (2nn!) has a limit.
I initially inferred that the question required me to use the definition of the limit of a sequence because a sequence is convergent if it has a limit |2nn!−L|<ϵ
I've come across approaches that use the squeeze theorem but I'm not sure whether its applicable to my question. While I have found answers on this site to similar questions containing the sequence, they all assume the limit is 0.
I think I need to show an≥an+1,∀n≥1, so
$$a_{n+1} = \frac{2^{n+1}}{(n+1)!}=\frac{2}{n+1}\frac{2^{n}}{n!}
A monotonic decreasing sequence is convergent and this particular sequence is bounded below by zero since its terms are postive. I'm not sure whether or not I need to do more to answer the question.
Answer
It is easy to prove that
∞∑k=12nn!<∞
e.g. with the ratio test you have
an+1an=n!2n⋅2n+1(n+1)!=2n+1⟶0
Then lim has to be 0
If you do not know anything about series you might assert, that n!>3^n if n\gt N_0 for some fixed N_0\in\mathbb{N}. Therefore you have for those n
0<\frac{2^n}{n!} \lt\frac{2^n}{3^n}=\left(\frac{2}{3}\right)^n\longrightarrow 0
Thus \lim\limits_{n\longrightarrow\infty} \frac{2^n}{n!} =0
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