- #1

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I rewrote the sum as [tex] \sum_{n=1}^{\infty} \frac{\sin(\frac{1}{n})}{\frac{1}{n}} [/tex]

Then I applied the Nth term test and used L'Hoptials rule so [tex] \lim_{n\to\infty} \frac{\cos(\frac{1}{n})\frac{-1}{n^2}}{\frac{-1}{n^2}} [/tex]

The [tex] \frac{-1}{n^2} [/tex] cancel out and the [tex] lim_{n\to\infty} \cos(\frac{1}{n}) [/tex] is 1 which by the nth term test is divergent. Is that a legitimate proof of divergence?