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

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 $$\epsilon$$?

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• Didn't notice that! I hope I will get lucky this weekend! – peastick Nov 16 '18 at 7:04

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.

• Thanks for the answer, but could you explain why your last inequality is true? And your first inequality should be $\leq$. – peastick Nov 16 '18 at 7:03
• @peastick Look the EDITED answer – Tito Eliatron Nov 16 '18 at 10:31