Evaluate limn→∞1n∑k=1ntan(kn2)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \tan(\frac{k}{n^2})limn→∞n1∑k=1ntan(n2k).
000
1/21/21/2
111
∞\infty∞