For large nnn, sin(1/n)≈1n−16n3+⋯\sin(1/n) \approx \frac{1}{n} - \frac{1}{6n^3} + \cdotssin(1/n)≈n1−6n31+⋯. By what series does ∑n=1∞sin(1/n)\sum_{n=1}^{\infty} \sin(1/n)∑n=1∞sin(1/n) behave?
∑(−1)n/n\sum (-1)^n/n∑(−1)n/n (conditionally convergent); the original series is conditionally convergent
∑1n\sum \frac{1}{n}∑n1 (diverges); the original series diverges
∑1n3\sum \frac{1}{n^3}∑n31 (converges); the original series converges
∑1n2\sum \frac{1}{n^2}∑n21 (converges); the original series converges