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

  • $\begingroup$ The sequence is bounded below and decreasing. Now assume it didn't converge to $0$. What can you say then about $\lim_{n\to\infty}n\left(\frac{b_n}{b_{n+1}}-1\right)$? $\endgroup$ – martin.koeberl Mar 29 '17 at 2:24
  • $\begingroup$ @martin.koeberl Then it has a nonzero and positive limit (which is impossible?). But I don't know how to prove it by absurd. $\endgroup$ – pjpj Mar 29 '17 at 2:26
  • $\begingroup$ I don't think it is trivial. If $b_n$ converges to a limit greater than zero, $\frac {b_n}{b_{n+1}}-1$ converges to zero. Certainly without the factor $n$ you couldn't claim that $b_n \to 0$ $\endgroup$ – Ross Millikan Mar 29 '17 at 2:45
  • $\begingroup$ Now, you have to use the definition of convergence in various ways. The keypoints being that above limit stays above $\lambda-\frac{\epsilon}2$ for any $\epsilon$ you choose (from some point), and that you can write $\frac{b_n}{b_{n+1}}-1=\frac{b_n-b_{n+1}}{b_{n+1}}$. For the numerator, use the definition of convergence, for the denominator, use that the limit is assumed to be strictly bigger than $0$. $\endgroup$ – martin.koeberl Mar 29 '17 at 2:46
  • $\begingroup$ @martin.koeberl Why is it above $\lambda-\frac{\epsilon}{2}$? $\endgroup$ – pjpj Mar 29 '17 at 3:10

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 $0$.


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.


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