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!
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.