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.

Limit of a sum

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