Find the Limit of the sequence $\frac{(2^n)(n!)}{(2n+1)!}$

Question

Find the Limit of the sequence $\frac{(2^n)(n!)}{(2n+1)!}$

I'm not sure how to approach this problem. I tried the squeeze method, but could not figure it out.

Answer 1

Note $$2^n\cdot n! = (2 \times \cdots \times 2) \times (n \times \cdots \times 1) = (2n) \times (2(n - 1)) \times \cdots \times 2.$$ Therefore $$\frac{2^n\cdot n!}{(2n + 1)!} = \frac{(2n) \times (2(n - 1)) \times \cdots \times 2}{(2n + 1) \times (2n) \times \cdots \times 1} = \frac{1}{(2n + 1)\times(2n - 1)\times\cdots\times 3 \times 1} \to 0$$ as $n \to \infty$.

Answer 2

We have

$$ a_n = \frac{2^nn!}{(2n+1)!} $$

So then we have

$$ a_n = \frac{2^nn!}{(2n+1)!} , a_{n+1} = \frac{2^{n+1}(n+1)!}{(2{(n+1)}+1)!} = \frac{2^{n+1}(n!)(n+1)}{(2n+3)!} $$

Taking the ratio test, $\frac{a_n}{a_{n+1}}$

$$ \frac{\frac{2^nn!}{(2n+1)!}}{\frac{2^{n+1}(n!)(n+1)}{(2n+3)!}}=\frac{2^n(n!)(2n+3)!}{2^{n+1}(2n+1)!(n!)(n+1)} =\frac{(2n+3)(2n+2)}{2(n+1)} =\frac{(2n+3)(2n+2)}{2n+2} = 2n+3 $$

So taking the limit:

$$ \lim_{n\rightarrow \infty} 2n+3 \rightarrow \infty $$

So the series converges. ($a_{n+1}$ becomes smaller and smaller compared to the preceding term.)

Answer 3

Moving to the n+1 term from this

$\frac{(2^n)(n!)}{(2n+1)!}$

You multiply by $\frac{2(n+1)}{(2n+2)(2n+3)}$ which result multiplication by:

$\frac{1}{2n+3}$

So the sequence converges to 0

Answer 4

Our guess for the limit of the sequence is $0$.

Consider the series $$\sum_{n=0}^{\infty} \frac{(2^n)(n!)}{(2n+1)!}$$

Apply the ratio test:

$$\left| \frac{a_{n+1}}{a_n} \right| = \frac{ \frac{(2^{n+1})(n+1)!}{(2n+3)!} }{ \frac{(2^n)(n!)}{(2n+1)!} }= \frac{2(n+1)}{(2n+3)(2n+2)}$$

Then,

$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{2(n+1)}{(2n+3)(2n+2)} = 0 < 1$$

Hence, the series converges.

A series $\sum a_n$ is convergent $\implies$ $\lim a_n = 0$