Show that $a_{k_0}\geq 2^{-k_0\epsilon}/C(\epsilon)$ for some $k_0$, when $\epsilon>0$ and $\sum_{k=1}^{\infty}a_k\geq 1$.

Question

Let $\epsilon >0$ and suppose that $\sum_{k=1}^{\infty}a_k\geq 1$. I need to show that there is $k_0$ and constant $C(\epsilon)$ such that $a_{k_0}\geq 2^{-k_0\epsilon}/C(\epsilon)$. I prove this by contradiction: assume that the inequality does not hold for any constant $C(\epsilon)$. Especially we can choose $C(\epsilon)=n$ so that $$ a_k<\frac{1}{n}2^{-k\epsilon} \quad \text{for all} \ n\in\mathbb{N} $$ But now using geometric series we have $$ \sum_{k=1}^{\infty}a_k<\frac{1}{n}\sum_{k=1}^{\infty}\bigg(\frac{1}{2^{\epsilon}}\bigg)^k<\frac{1}{n}\frac{1}{2^{\epsilon}-1}<1 $$ for $n$ large enough. This is a contradiction.

My question is: Is my proof correct and why the constant $C$ has to be dependent on $\epsilon$?

Answer 1

Your original statement is $\forall \varepsilon>0$, $\exists C>0$ and $k_0\in{\mathbb N}$ such that $a_{k_0}\ge\frac{1}{C}2^{-k_0\varepsilon}$.

So if the negation should be something like:

$\exists \varepsilon_0>0$ such that for any $C>0$ and any $k\in{\mathbb N}$, it holds that $a_k<\frac{1}{C}2^{-k\varepsilon_0}$.

For THAT $\varepsilon_0>0$,then you can use directly the last inequality for $C=1$, Hence, $$1<\sum_{k=1}^\infty a_k \le \sum_{k=1}^\infty 2^{-k\varepsilon_0} =\frac{2^{-\varepsilon_0}}{1-2^{-\varepsilon_0}}< 1,$$ which is a contradiction.

So, your argument is almost correct, but you have to polish some steps.

EDIT Maybe some steps are not clear or wrong. Here is a better argument:

For that $\varepsilon_0$, take $C=\frac{1-2^{-\varepsilon_0}}{2^{1-\varepsilon_0}}>0$, so $$1\le \sum_{k=1}^\infty a_k \le C \sum_{k=1}^\infty 2^{-k\varepsilon_0} =C\frac{2^{-\varepsilon_0}}{1-2^{-\varepsilon_0}}=\frac{1}{2},$$ which is a contradiction.