Fast convergence in $L^1$ implies convergence almost everywhere
Yes, your proof is correct.
Moreover, instead of $\frac{1}{n^p}$ there could have been any function $f$, such that $\Sigma_{i = 1}^\infty f(n)$ converges.
Fast convergence in $L^1$ implies convergence almost everywhere
Yes, your proof is correct.
Moreover, instead of $\frac{1}{n^p}$ there could have been any function $f$, such that $\Sigma_{i = 1}^\infty f(n)$ converges.