limits - 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 $\lim_{n\rightarrow\infty}b_n=0$?

Answer

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 $0$.