Assume $\sum a_k$ is a series of positive terms that converges. Prove that $\sum \frac1{a_k}$ is divergent

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\sum \frac{1}{a_{k}}$.

Answer 4

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 $1/a_k\not\to 0$