Prove that the alternating series converges. 
Given that
  $\lim_{n\rightarrow\infty}n\left(\frac{b_n}{b_{n+1}}-1\right)=\lambda>0$,
  show that
  $\sum_{n=1}^{\infty}{\left(-1\right)^{n}b_{n}}\left(b_{n}>0\right) $
  converges.

Using the definition of limit of sequence, I can prove that $\left\{b_n\right\}$ is monotonically decreasing when $n$ is large enough. But how to prove $\lim_{n\rightarrow\infty}b_n=0$?
 A: Given $0 < r < \lambda$, there exists $N$ such that if $n \geqslant N$ we have
$$n \left(\frac{b_n}{b_{n+1}}-1  \right)> r \\ \implies \frac{b_n}{b_{n+1}} > 1 + \frac{r}{n}.$$
Hence for all $m > N$ it follows that
$$\frac{b_N}{b_m} > \prod_{k=N}^{m-1}\left(1 + \frac{r}{k} \right).$$
The infinite product on the RHS diverges to $+ \infty$ as $ m \to \infty$ since $\sum 1/k $ diverges. Therefore,  $b_m$ converges to $0$.
A: If
$\lim_{n\rightarrow\infty}n\left(\frac{b_n}{b_{n+1}}-1\right)
=c>0
$,
then,
arguing loosely,
$\frac{b_n}{b_{n+1}}-1
\approx c$
so
$\frac{b_n}{b_{n+1}}
\approx 1+c$.
Therefore,
for any $0 < d < c$,
for all large enough $n$,
$\frac{b_n}{b_{n+1}}
\ge 1+d$.
By induction,
$\frac{b_n}{b_{n+k}}
\ge (1+d)^k$
or
$b_{n+k}
\le \frac{b_n}{(1+d)^k}
\to 0$
as $k \to \infty$.
Therefore
$\lim_{n \to \infty} b_n = 0$.
Since
$\frac{b_{n+1}}{b_n}
\le \frac1{1+d}$
for all large enough $n$,
$b_n$ is a decreasing sequence
for large enough $n$.
By the alternating series theorem
(alternating sign,
limit zero, decreasing),
the sum converges.
