Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $ {t_n}$ be sequence on positive numbers how prove this series $$\sum_{n=1}^\infty\left(\frac1n\right)\cdot \frac{t_{n+1}+1}{t_n} $$ is divergent thanks in advance

share|cite|improve this question
up vote 4 down vote accepted

Let $a_n = \frac1n \frac{t_{n+1}+1}{t_n}$ be the series in question. If $t_n$ is bounded then $a_n \ge 1/(n \sup t_n)$ so $\sum a_n$ clearly diverges.

On the other hand, if $t_n$ is unbounded then we have an increasing subsequence $t_{n_1} < t_{n_2} < t_{n_3} < \cdots$. By passing to a thinner subsequence we may assume WLOG that $n_{i+1} \ge 2n_i$ for each $i$. We now estimate the sum $\sum_{n_i \le k < n_{i+1}} a_k$ using AM-GM. Let $d_i = n_{i+1}-n_i$ be the number of terms in this sum. Note that $$\prod_{n_i \le k < n_{i+1}} a_k > \prod_{n_i \le k < n_{i+1}} \frac1{n_{i+1}} \frac{t_{k+1}}{t_k} = \big(\frac1{n_{i+1}}\big)^{d_i} \frac{t_{n_{i+1}}}{t_{n_i}} > \big(\frac1{n_{i+1}}\big)^{d_i},$$

so the arithmetic/geometric mean inequality gives

$$\sum_{n_i \le k < n_{i+1}} a_k \ge d_i \big(\frac1{n_{i+1}}\big) \ge \frac12.$$

By summing this over all $i$ we see that $\sum a_n$ diverges.

share|cite|improve this answer
thanks that helped. – Maisam Hedyelloo Jan 17 '13 at 22:59
@MaisamHedyelloo If this answers your question, please accept it. See this page on how to do so: – Ayman Hourieh Jan 18 '13 at 1:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.