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Limit of a sum

Let a_n be a real sequence such that a_n>0 , \liminf a_n=1 , \limsup a_n =2 and \lim \limits_{n \rightarrow +\infty} \sqrt[n] {\prod \limits_{k=1}^n{a_k}}=1. Prove that

    \[\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} a_k =1\]

Solution

Lemma: Let a_n be a bounded sequence of, say, complex numbers and let \ell be another complex number.

    \[\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k=\ell \Leftrightarrow \text{D-lim } a_n=\ell\]

where the latter means “convergence on a set of density 1 ” i.e. there exists a subsequence n_1,n_2,\dotsc such that \lim \limits_{k \to +\infty} a_{n_k}=\ell  and n_k is dense i.e.

    \[\lim \limits_{n \to +\infty} \frac{\vert \{1,2,\dotsc,n\} \cap \{n_1,n_2,\dotsc,\}\vert}{n}=1\]

Proof: The proof is omitted because it is too technical.

Hence,

    \begin{align*} \lim_{n \rightarrow +\infty} \sqrt[n]{\prod_{k=1}^n a_k}=1 &\Leftrightarrow \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^n \log a_k=0 \\ &\Leftrightarrow \text{D-lim } \log a_n=0 \\ &\Leftrightarrow \text{D-lim } a_n=1 \\ &\Leftrightarrow \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^n a_k=0 \end{align*}

All we need is that the logarithm is continuous and that a_n as well as \log a_n are bounded sequences.

Note: We can simplify the conditions to a_n>0, \liminf \limits_{n \rightarrow +\infty} a_n>0 and \limsup \limits_{n \rightarrow +\infty} a_n< +\infty.

 

The exercise along with the solution may be found on AoPS.com . The proof of the claim may also be found there.

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