Let \(a_1,a_2,\ldots\) be positive real numbers.
Prove that \[ \sum_{n=1}^{\infty}a_n<\infty \quad\Longrightarrow\quad \sum_{n=1}^{\infty}\sqrt{a_na_{n+1}}<\infty. \]
Prove that the converse is false.
Solution
1. For every \(n\ge1\),\[ \sqrt{a_na_{n+1}}\le\frac{a_n+a_{n+1}}2. \]
This is the arithmetic-geometric mean inequality for the positive numbers \(a_n\) and \(a_{n+1}\).
2. If\[ \sum_{n=1}^\infty a_n<\infty, \]then\[ \sum_{n=1}^\infty\sqrt{a_na_{n+1}}<\infty. \]
By step 1, for every \(N\), \[ \begin{aligned} \sum_{n=1}^N\sqrt{a_na_{n+1}} &\le\frac12\sum_{n=1}^N(a_n+a_{n+1})\\ &\le\sum_{n=1}^{N+1}a_n. \end{aligned} \] The right-hand side is bounded uniformly in \(N\) by the convergent series \(\sum a_n\). Since the partial sums on the left are increasing, they converge.
3. Define a positive sequence by\[ a_{2k-1}=\frac1{k^3}, \qquad a_{2k}=\frac1k \qquad(k\ge1). \]Then\[ \sum_{n=1}^\infty a_n=\infty. \]
The even-indexed subseries is \[ \sum_{k=1}^\infty a_{2k} = \sum_{k=1}^\infty\frac1k, \] which diverges.
4. For the sequence in step 3,\[ \sum_{n=1}^\infty\sqrt{a_na_{n+1}}<\infty. \]
For \(k\ge1\), \[ \sqrt{a_{2k-1}a_{2k}} = \frac1{k^2}. \] Also \[ \sqrt{a_{2k}a_{2k+1}} = \frac{1}{\sqrt{k(k+1)^3}} \le \frac1{k^2}, \] because \(k(k+1)^3\ge k^4\). Hence \[ \sum_{n=1}^\infty\sqrt{a_na_{n+1}} \le 2\sum_{k=1}^\infty\frac1{k^2} <\infty. \]
5. Thus the converse is false.
Steps 3 and 4 give a positive sequence for which the geometric-mean series converges while \(\sum a_n\) diverges.
6. Q.E.D.
Step 2 proves part 1, and step 5 proves part 2.