Assume $\sum a_k$ is a series of positive terms that converges. Prove that $\sum \frac1{a_k}$ is divergent I started this problem by looking that $a_k \to 0$ for the limit of partial sums to converge. So then $\frac {1}{a_k} \nrightarrow 0$. I feel like I am missing something though to make this concrete.
 A: Yes you are right: $\lim_{k\to\infty}a_k =0$ but this implies  $\lim_{k\to\infty}\frac{1}{a_k} =\infty$ and summing up unbounded elements will render a divergent sum.
A: If $\sum a_k$ is a convergent sum  and
$\sum b_k$ is is a convergent sum  and $1 = a_k b_k$ for all $k$, then $a_k$ and $b_k$ both have a limit of $0.$ Then taking the limit of their products gives $1 = 0 \cdot 0 = 0$ which is false. The result you want to prove follows.
A: For simplicity assume that:
$$ P\varepsilon \ge a_{k} \ge \varepsilon$$
So:
$$ \sum_{k=1} ^{M} P\varepsilon \ge \sum_{k=1} ^{M} a_{k} \ge \sum_{k=1} ^{M} \varepsilon \Rightarrow P\varepsilon \sum_{k=1} ^{M} 1  \ge \sum_{k=1} ^{M} a_{k} \ge \varepsilon \sum_{k=1} ^{M} 1 $$
$$ \Rightarrow M P\varepsilon \ge \sum_{k=1} ^{M} a_{k} \ge M\varepsilon $$
For convergence of $a_{k}$ at $ \lim_{M\to\infty}$ we must have a limited to zero epsilon:
$$\Rightarrow \lim_{M\to\infty}\varepsilon=0 $$.
By writing this for $\sum \frac{1}{a_{k}}$ :
$$ \frac{1}{P\varepsilon} \le \frac{1}{a_{k}} \le \frac{1}{\varepsilon}$$
$$ \sum_{k=1} ^{M} \frac{1}{P\varepsilon} \le \sum_{k=1} ^{M} \frac{1}{a_{k}} \le \sum_{k=1} ^{M} \frac{1}{\varepsilon} \Rightarrow \frac{1}{P\varepsilon} \sum_{k=1} ^{M} 1  \le \sum_{k=1} ^{M} \frac{1}{a_{k}} \le \frac{1}{\varepsilon} \sum_{k=1} ^{M} 1 $$
$$ \Rightarrow \frac{M}{P\varepsilon} \le \sum_{k=1} ^{M} \frac{1}{a_{k}} \le \frac{M}{\varepsilon} $$
By $\lim_{M\to\infty}\varepsilon=0 $  we have divergence for $\sum \frac{1}{a_{k}}$.
A: Since the series is convergent $a_k\to 0$. In particular there exists $N$ such that $0< a_k\leq 1/2$ for $k\geq N$. In particular for $k\geq N$, it follows that $1/a_k\geq 2$ which implies that $1/a_k\not\to 0$
