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But, along the subsequence of powers of $2$, we have $a_n n \log (n) = \log (n) \to \infty$ To do this we observe that $\lim_ {x\to 0}\frac {\vert ln (1+x)\vert} {\vert x\vert}=1$ by the l'hopital rule. Erdős called this 'perhaps my first serious problem' (in [er98] he dates it to 1931)

The powers of $2$ show that $2^n$ would be best possible here Taking the logarithm of the product gives the series… The trivial lower bound is $n \gg 2^ {n}/n$, since all $2^n$ distinct subset sums must lie in $ [0,nn)$.

In words, if \ (\omega \in a_n\) does not hold for infinitely many \ (n\), then we can infer two things

\ (\omega \in a_n^c\) for all but finitely many \ (n\) Hence \ [p (limsup a_n) = 0 \implies p (limsup a_n^c) = 1 \implies p (liminf a_n^c) = 1\] which one we end up using depends on the problem. Definition (not explicitly in text) A sequence an a n diverges to −∞ ∞ if and only if for any K> 0 K> 0, there exists n∗ ∈N n ∗ ∈ N such that an <−K a n <K for all n ≥ n∗ n ≥ n ∗. If this is the case, we say that the limit exists and we write limn→∞an = −∞ lim n → ∞ a n = ∞. There is also a correct way to 'reverse' the statement in your claim, but this is a syntactic reversal, creating a second statement that is logically equivalent to the first.

Let $latex a_n$ be a sequence of positive numbers Then the infinite product $latex \displaystyle\prod_ {n=1}^ {\infty} (1+a_n)$ converges if and only if the series $latex \displaystyle\sum_ {n=1}^ {\infty} a_n$ converges

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