We've only been taught to find limits using the Squeeze Theorem and L'Hopitals Rule, so I'm not sure how to go about finding the limit of this sequence.
Answer
For a simpler idea, note that $$\frac{a_{n+1}}{a_n}=\frac{n+1}2$$
It follows that $a_{n+1}/a_n>2$ for $n=3,4,\dots$ so that $a_n$ grows faster than $2^n$; thus it diverges to $+\infty$.
Hint Set $$a_n=\frac{n!}{ 2^n}$$ and show that $a_n$ is strictly increasing for $n\geq 3$. That is, look at when $${a_{n+1}}>{a_n}$$ is true.
$2^n$ is tiny compared to $n!$ since we're multiplying by a constant factor of $2$, while in $n!$ we're constantly increasing the factor by $1$.
LEMMA $$n!/2^{n-1}$$ is a (positive) integer for infinitely many values of $n$.
P Take $n=2^k$ for $k=0,1,2,\dots$. The multiplicity for which $2$ divides $2^k!$ is $$v_2(2^k)=\sum_{m=1}^\infty\left\lfloor \frac{2^k}{2^m}\right\rfloor\\=\sum_{m=1}^\infty\lfloor 2^{k-m}\rfloor\\=\sum_{m=1}^k 2^{k-m}=1+2+\cdots+2^{k-1}=2^k-1$$
which means $2^{2^k-1}\mid 2^k!$.
SPOILER
You should be able to show that $$a_n\to +\infty$$
No comments:
Post a Comment