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Pietro Majer
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Since $\sum_n a_n/b_n < \infty$$\sum_ {n=1}^\infty \frac{a_n}{b_n } < \infty$ and $a_n/b_n \ge0$$0 \le \frac{a_n}{b_n + \sigma/N}\le \frac{a_n}{b_n} $, it follows we have that $f_N(\sigma)\to \sum_ {n=1}^\infty a_n/b_n$$\sum_{n=1}^N \frac{a_n}{b_n + \sigma/N} \to \sum_ {n=1}^\infty \frac{a_n}{b_n }$ as $N\to\infty$, just by dominated convergence.

Since $\sum_n a_n/b_n < \infty$ and $a_n/b_n \ge0$, it follows that $f_N(\sigma)\to \sum_ {n=1}^\infty a_n/b_n$ just by dominated convergence.

Since $\sum_ {n=1}^\infty \frac{a_n}{b_n } < \infty$ and $0 \le \frac{a_n}{b_n + \sigma/N}\le \frac{a_n}{b_n} $, we have that $\sum_{n=1}^N \frac{a_n}{b_n + \sigma/N} \to \sum_ {n=1}^\infty \frac{a_n}{b_n }$ as $N\to\infty$, just by dominated convergence.

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Pietro Majer
  • 60.6k
  • 4
  • 122
  • 269

Since $\sum_n a_n/b_n < \infty$ and $a_n/b_n \ge0$, it follows that $f_N(\sigma)\to \sum_ {n=1}^\infty a_n/b_n$ just by dominated convergence.