Does there exist a divergent series \(\sum_{i=0}^\infty a_i\) of positive real numbers \(a_i\) such that \(\sum_{i=0}^\infty \sqrt{a_i}\) converges? If so, give an example; if not, prove it.
<1>2. Suppose that \(\sum_{i=0}^\infty \sqrt{a_i}\) converges. Then there is an index \(N\) such that \(0<a_i<1\) for every \(i\ge N\).
<1>3. For every \(i\ge N\), \[ a_i\le \sqrt{a_i}. \]
<1>4. The series \(\sum_{i=0}^\infty a_i\) converges.
<1>5. Therefore it is impossible for \(\sum a_i\) to diverge while \(\sum\sqrt{a_i}\) converges.
Solution
1. No such series exists.
Convergence of \(\sum \sqrt{a_i}\) implies \(\sqrt{a_i}\to0\), hence \(a_i\to0\). Therefore \(a_i<1\) eventually.
If \(0<a_i<1\), then squaring decreases the number: \((\sqrt{a_i})^2=a_i\le\sqrt{a_i}\).
By 3, \[ 0\le \sum_{i=N}^m a_i\le \sum_{i=N}^m \sqrt{a_i} \] for every \(m\ge N\). The right-hand partial sums are bounded because \(\sum \sqrt{a_i}\) converges. Hence the positive-term series \(\sum_{i=N}^\infty a_i\) converges by comparison. Adding the finite initial sum \(\sum_{i=0}^{N-1}a_i\) preserves convergence.