Assume \(\beta >0\), \(a_n>0\), \(n=1,2,\ldots\), and the series \(\sum a_n\) is divergent. Show that \(\displaystyle \sum \frac{a_n}{\beta + a_n}\) is also divergent.
Solution
1. Claim: \(\sum \frac{a_n}{\beta + a_n}\) diverges.
we show its partial sums are unbounded, using the divergence of \(\sum a_n\).
2. Two-case bound: if \(a_n \le \beta\) then \(\frac{a_n}{\beta + a_n} \ge \frac{a_n}{2\beta}\); if \(a_n > \beta\) then \(\frac{a_n}{\beta + a_n} > \frac{1}{2}\).
\(a_n \le \beta \Rightarrow \beta + a_n \le 2\beta\); \(a_n > \beta \Rightarrow \beta + a_n < 2a_n \Rightarrow \frac{a_n}{\beta + a_n} > \frac{1}{2}\).
3. Hence \(\frac{a_n}{\beta + a_n} \ge \frac{1}{2}\min\left\{\frac{a_n}{\beta}, 1\right\}\) for every \(n\).
2 covers both cases uniformly: for \(a_n \le \beta\) the min is \(a_n/\beta\) and the RHS is \(a_n/(2\beta)\); for \(a_n > \beta\) the min is \(1\) and the RHS is \(1/2\).
4. \(\sum \min\left\{\frac{a_n}{\beta}, 1\right\}\) diverges.
let \(I = \{n : a_n \ge \beta\}\).
If \(\sum_{n \in I} 1 = \infty\) we are done. Otherwise \(I\) is finite, so for all but finitely many \(n\), \(\min\{a_n/\beta, 1\} = a_n/\beta\), and \(\sum a_n/\beta = \infty\) (as \(\sum a_n\) diverges). Either way the sum diverges.
5. Q.E.D.
3 and 4 give \(\sum \frac{a_n}{\beta + a_n} \ge \frac12 \sum \min\{a_n/\beta, 1\} = \infty\).