Compute the following limits:
- \(\lim_{n\to\infty} \sum_{k\geq 1} {1\over k^2} \sin^n(k)\)
- \(\lim_{n\to\infty} \sum_{k\geq 1} {1\over k} e^{-k/n}\)
Solution
Both sums are integrals against counting measure on \(\theset{1, 2, \ldots}\).
1. \(\lim_{n\to\infty} \sum_{k\geq 1} \frac{\sin^n(k)}{k^2} = \boxed{0}\).
1.1. \(\abs{\sin k} < 1\) for every integer \(k \geq 1\).
\(\abs{\sin k} = 1\) would force \(k = \pi/2 + m\pi\) for some \(m \in \ZZ\), so \(\pi = 2k/(2m+1)\) would be rational.
1.2. \(\sin^n(k)/k^2 \to 0\) for each \(k\), and \(\abs{\sin^n(k)/k^2} \leq 1/k^2\).
The limit follows from step 1, and the bound from \(\abs{\sin k} \leq 1\).
1.3. Q.E.D.
\(\sum_k 1/k^2 < \infty\), so by step 2 the dominated convergence theorem gives \(\lim_n \sum_k \sin^n(k)/k^2 = \sum_k 0 = 0\).
2. \(\lim_{n\to\infty} \sum_{k\geq 1} \frac{e^{-k/n}}{k} = \boxed{\infty}\).
The terms are nonnegative and \(e^{-k/n}/k \to 1/k\) for each \(k\). By Fatou’s lemma, \[ \liminf_{n\to\infty} \sum_{k\geq 1} \frac{e^{-k/n}}{k} \geq \sum_{k\geq 1} \liminf_{n\to\infty} \frac{e^{-k/n}}{k} = \sum_{k\geq 1} \frac{1}{k} = \infty. \]