An article from AMM vol 95 Page 942. (I have some doubt about it) Today I am reading an article in American Math Monthly volume 95 Page 942.
Author introduces an alternating series:$$\sum_{n=1}^{\infty}(-1)^n\frac{(2n)!}{4^n(n!)^2}$$
then he uses the Stirling's formula and alternating series test to conclude that the above series is conditional convergent. Using Stirling's formula, we have $$\frac{(2n)!}{4^n(n!)^2} \sim \frac{1}{\sqrt{n}}$$ and I think this facts implies that $\frac{(2n)!}{4^n(n!)^2} \rightarrow 0$ as $n \rightarrow \infty$. But the author says that it is a little more difficult to obtain $\frac{(2n)!}{4^n(n!)^2} \rightarrow 0$ and he introduces Sapagof's Test : 
Theorem

Suppose$\{a_n\}$ is a decreasing sequence of positive numbers and for each natural number $n$, define $b_n = 1-a_{n+1}/a_n$. Then the sequence$\{a_n\}$converges to zero if and only if the series $\sum b_n$ diverges.

He uses the above theorem to conclude that $a_n \rightarrow 0$. What's wrong with me? Thanks very much.
 A: Out of doubt if a sequence $(a_n)$ verify $a_n\sim_\infty b_n$ and $\displaystyle\lim_{n\to\infty}b_n=0$ then $\displaystyle\lim_{n\to\infty}a_n=0$. In fact, this result is clear from  the definition of $a_n\sim b_n$: there's a sequence $(v_n)$ such that $\displaystyle\lim_{n\to\infty}v_n=1$ and there's $n_0\in \mathbb N$:
$$\forall n\geq n_0,\quad a_n=v_nb_n$$
but even if the series $\sum_n b_n$ is convergent we can not conclude directly that the series $\sum_n a_n$ is also convergent unless $a_n$ has a positive sign. For example if $a_n=\frac{(-1)^n}{\sqrt{n}+(-1)^n}$ then $a_n\sim_\infty b_n=\frac{(-1)^n}{\sqrt{n}}$ and the series $\sum_n b_n$ is convergent while the series $\sum_n a_n$ is divergent.
A: We are looking at
$\sum_{n=1}^{\infty}(-1)^n\dfrac{(2n)!}{4^n(n!)^2}$
(I find dfrac easier to read than frac).
If $a_n 
= \dfrac{(2n)!}{4^n(n!)^2}
$,
$
\begin{align}
\dfrac{a_{n+1}}{a_n}
&= \dfrac{(2(n+1))!}{4^{n+1}((n+1)!)^2}\big/\dfrac{(2n)!}{4^n(n!)^2}\\
&= \dfrac{(2(n+1))!4^n(n!)^2}{4^{n+1}((n+1)!)^2(2n)!}\\
&= \dfrac{(2n+1)(2n+2)}{4(n+1)^2}\\
&= \dfrac{2n+1}{2(n+1)}\\
&= 1-\dfrac{1}{2(n+1)}\\
\end{align}
$
so
$a_n$ is strictly decreasing.
Since $\sum \dfrac{1}{2(n+1)} $
 diverges,
$a_n \to 0$.
By the alternating series test,
$\sum (-1)^n a_n$ converges.
Also,
$\dfrac{a_{n+1}}{a_n}
=1-\dfrac{1}{2(n+1)}
>1-\dfrac{1}{n+1}
=\dfrac{n}{n+1}
$,
so
$\dfrac{a_{N}}{a_M} 
=\prod_{n=M}^{N-1} \dfrac{a_{n+1}}{a_n}
>\prod_{j=M}^{N-1} \dfrac{n}{n+1}
=\dfrac{M}{N}
$.
Therefore,
since $\sum \dfrac{1}{N}$ diverges,
$\sum a_n$ diverges.
A: It's easy to see that
$$\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}=\frac{1}{1+\dfrac{1}{2n+1}},\ \forall n\in\mathbb{N}.$$
So, by Bernoulli's inequality,
$$\frac{a_n}{a_1}=\prod_{k=1}^{n-1}\frac{a_{k+1}}{a_k}=\prod_{k=1}^{n-1}\frac{1}{1+\dfrac{1}{2k+1}}<\frac{1}{\sum\limits_{k=1}^{n}\dfrac{1}{2k-1}}\to 0\ (n\to\infty).$$
This implies
$$\lim_{n\to\infty}a_n=0.$$
