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Let $ (a_k)_{k\in\mathbb{N}}$ be a decreasing sequence of positive real numbers. We suppose that there exists a $b>0$ such that $a_k \geq \frac{b}{k}$ for infinite values of $k$ . Prove that the series $\sum_{k=1}^{\infty} a_k$ is divergent.

Can anybody help me with this calculus exercise? Thanks in advance!

dmvlt
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    compare to the harmonic series $\sum \frac1n$ – Gregory Grant May 04 '15 at 16:15
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    Agree, but the problem is that the inequality holds for infinite values of $k$, not for all $k \in \mathbb{N}$ . – dmvlt May 04 '15 at 16:20
  • Right, that's where "decreasing" comes into play. – Gregory Grant May 04 '15 at 16:22
  • We can "choose" the subsequence $ (a_{n_k})$ which satisfies the inequality $a_{n_k} \geq \frac{b}{n_k} , \forall k \in \mathbb{N} $ and use the monotonicity: $n_k \geq k \Rightarrow a_{n_k} \leq a_k $ , for all $k \in\mathbb{N}$. So, $\frac{b}{n_k} \leq a_{n_k} \leq a_k , \forall k \in\mathbb{N} $ . But, can we assume that the series $\sum_{k=1}^{\infty} \frac{b}{n_k}$ diverges? – dmvlt May 04 '15 at 17:27
  • Yeah, that's where I get stuck. – dmvlt May 04 '15 at 17:36
  • I think I got it, please check my answer below and let me know if you need more details. – Gregory Grant May 04 '15 at 18:16

1 Answers1

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Choose $k_1$ such that $a_{k_1}\geq\frac{b}{k_1}$. Then $\sum_{j=1}^{k_1}a_j\geq k_1a_{k_1}\geq b$.
Now for $i\geq1$ choose $k_{i+1}$ such that $a_{k_{i+1}}\geq\frac{b}{k_{i+1}}$ and $k_{i+1}\geq\frac{i+1}{i}k_i$. You can do this because there are infinitely many $k$ satisfying the condition. Then $\sum_{j=k_i+1}^{k_{i+1}}a_j\geq(k_{i+1}-k_i)a_{k_{i+1}}\geq(k_{i+1}-k_i)\frac{b}{k_{i+1}}\geq\frac{b}{i+1}$. Thus we have rewritten the sum $\sum a_k$ as a sum $\sum c_i$ where each $c_i\geq\frac{b}{i+1}$. Thus $\sum c_i$ diverges. Thus $\sum a_k$ diverges.

Gregory Grant
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  • Very good approach. But why $(k_{i+1} - k_i ) \frac{b}{k_{i+1}} \geq \frac{b}{i+1} $ ? From the inequality $k_{i+1} \geq \frac{i+1}{i} k_i $ we have that $ \frac{1}{i+1} \geq \frac{k_i}{i k_{i+1}} $. – dmvlt May 04 '15 at 23:52
  • $(k_{i+1}-k_i)\frac{b}{k_{i+1}}\geq\frac{b}{i+1}$ $\Longleftrightarrow$ $(i+1)(k_{i+1}-k_i)\geq k_{k+1}$ $\Longleftrightarrow$ $ik_{i+1}+k_{i+1}-(i+1)k_i\geq k_{i+1}$ $\Longleftrightarrow$ $ik_{i+1}\geq(i+1)k_i$ $\Longleftrightarrow$ $k_{i+1}\geq\frac{i+1}{i}k_i$. That proves it. – Gregory Grant May 04 '15 at 23:58
  • You're welcome, it was an interesting exercise, harder than it seemed at first glance. – Gregory Grant May 05 '15 at 00:02