Mathematics III - Foundations of Analysis - Summer 24

Alternatively we could choose $N_1$ and $N_2$ as in my post from 13:35 but then we need to take $N=\max\{2N_1,2N_2+1\}.$ Just $\max\{N_1,N_2\}$ here might be too small.

Then if $n>N$ and $n=2k$ we get $|x_{2k}-L| < \varepsilon$ since $n=2k>N\geq 2N_1$ so that $k>N_1$, and if instead $n=2k+1$ we still get $|x_{2k+1}-L| < \varepsilon$ since $n=2k+1>N\geq 2N_2+1$ so $k>N_2$, so in both cases we get $|x_n - L| < \varepsilon$.