Lindelöf summation

Remarks

Example: let $\sum a_n$ converge to $s$. For fixed $h > 0$ the factors $n^{-hn} = e^{-hn \ln n}$ decrease in $n \ge 1$ and lie in $(0, 1]$. By Abel's partial summation, $L(h)$ converges, and the tails of $L(h)$ are bounded by the tails of $\sum a_n$ uniformly in $h$. Since each factor tends to $1$ as $h \to 0^+$, one gets $L(h) \to s$, so the series is Lindelöf summable to its sum.

Non-example: for $1 + 1 + 1 + \cdots$, $L(h) = 1 + \sum_{n \ge 1} n^{-hn}$ is defined for every $h > 0$. For each $N$, $L(h) \ge 1 + \sum_{n=1}^{N} n^{-hn} \to N + 1$ as $h \to 0^+$. So $L(h) \to +\infty$, and the second condition fails.