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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

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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.

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thanks that helped. –  Maisam Hedyelloo Jan 17 '13 at 22:59
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@MaisamHedyelloo If this answers your question, please accept it. See this page on how to do so: meta.stackoverflow.com/a/5235 –  Ayman Hourieh Jan 18 '13 at 1:22
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