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Let $(a_n)$ positive sequence where $$\dfrac{n-1}{n} \leq \dfrac{a_{n+1}}{a_n} \leq \dfrac {n}{n+1}$$

Show that $\sum_{n=1}^{\infty} a_n$ diverges.

I already found out that $\sum_{n=1}^{\infty} (-1)^na_n$ is convergent and that $\lim a_n =0$.

I'm not sure how to approach this one.

1 Answers 1

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For all $n \ge 1$,

$$a_{n+1} \ge \frac{n-1}{n}a_n \ge \frac{(n-1)(n-2)}{n(n-1)}a_{n-1}\ge \cdots \ge \frac{(n-1)(n-2)\cdots 2\cdot 1}{n(n-1)\cdots 3\cdot 2}a_2 = \frac{1}{n}a_2$$

Since $\sum \frac{1}{n}a_2$ diverges, by comparison the series $\sum a_n$ diverges.

  • 0
    Can you explain why $\sum \dfrac{1}{n}a_2$ diverges? I can use that $\sum \dfrac{1}{n}$ diverges, but does it means that also the first one? – 2017-01-10
  • 1
    Let $s_n$ be the sequence of partial sums of $\sum \frac{1}{n}$. Then $s_n$ is divergent, so then $a_2 s_n$ is divergent. But $a_2 s_n$ is the sequence of partial sums of $\sum \frac{1}{n}a_2$. So $\sum \frac{1}{n} a_2$ diverges. In general, if $\sum c_n$ diverges, then so does $\sum ac_n$ for any nonzero constant $a$. – 2017-01-10