Define what is meant for an infinite series \(\sum_{n=1}^\infty b_n\) of real numbers \(b_n\) to converge.
Let \(a_1, a_2, \dots\) be a sequence of positive real numbers such that \(a_1 > a_2 > \cdots > a_n > a_{n+1} > \cdots\) and \(\lim_{n \to \infty} a_n = 0\). Prove that the infinite series \(\sum_{n=1}^\infty (-1)^n a_n\) converges.
Solution
The series \(\sum_{n=1}^\infty b_n\) converges if its sequence of partial sums \[ s_N=\sum_{n=1}^N b_n \] converges to a finite real number.
It is enough to prove convergence of \[ t_N=\sum_{n=1}^N(-1)^{n+1}a_n, \] because the series in the question has partial sums \(-t_N\). For even indices, \[ t_{2m}=(a_1-a_2)+(a_3-a_4)+\cdots +(a_{2m-1}-a_{2m}), \] so \((t_{2m})\) is increasing. Also \(t_{2m}<a_1\), since \[ t_{2m}=a_1-(a_2-a_3)-\cdots-(a_{2m-2}-a_{2m-1})-a_{2m}<a_1. \] Hence \(t_{2m}\to L\) for some \(L\).
Moreover \[ t_{2m+1}-t_{2m}=a_{2m+1}\to0, \] so \(t_{2m+1}\to L\) as well. Therefore the full sequence \((t_N)\) converges to \(L\), and consequently \[ \sum_{n=1}^\infty(-1)^n a_n \] converges to \(-L\).