1) Examine sum_{n=1}^{infty} frac{operatorname{tg} frac{1}{n^{2}}}{n+3} for convergence

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Step:1 ：(S = sum_{n=1}^infty frac{operatorname{tg} frac{1}{n^2}}{n+3})Step:2 ：Compare with (sum_{n=1}^infty frac{1}{n^2}), which converges by the p-series test with p=2Step:3 ：Use the comparison test: (0 leq frac{operatorname{tg} frac{1}{n^2}}{n+3} leq frac{1}{n^2}) for all nStep:4 ：Since (sum_{n=1}^infty frac{1}{n^2}) converges, so does (sum_{n=1}^infty frac{operatorname{tg} frac{1}{n^2}}{n+3}) by the comparison testStep:5 ：(boxed{text{The series converges}})

Answer

Step: 1 ：(S = sum_{n=1}^infty frac{operatorname{tg} frac{1}{n^2}}{n+3})Step: 2 ：Compare with (sum_{n=1}^infty frac{1}{n^2}), which converges by the p-series test with p=2Step: 3 ：Use the comparison test: (0 leq frac{operatorname{tg} frac{1}{n^2}}{n+3} leq frac{1}{n^2}) for all nStep: 4 ：Since (sum_{n=1}^infty frac{1}{n^2}) converges, so does (sum_{n=1}^infty frac{operatorname{tg} frac{1}{n^2}}{n+3}) by the comparison testStep: 5 ：(boxed{text{The series converges}})

1) Examine $\sum_{n=1}^{\infty} \frac{\operatorname{tg} \frac{1}{n^{2}}}{n+3}$ for convergence

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