chimychang
Mar6-10, 10:48 PM
\sum_{n=1}^{\infty} n \sin(\frac{1}{n})
I rewrote the sum as \sum_{n=1}^{\infty} \frac{\sin(\frac{1}{n})}{\frac{1}{n}}
Then I applied the Nth term test and used L'Hoptials rule so \lim_{n\to\infty} \frac{\cos(\frac{1}{n})\frac{-1}{n^2}}{\frac{-1}{n^2}}
The \frac{-1}{n^2} cancel out and the lim_{n\to\infty} \cos(\frac{1}{n}) is 1 which by the nth term test is divergent. Is that a legitimate proof of divergence?
I rewrote the sum as \sum_{n=1}^{\infty} \frac{\sin(\frac{1}{n})}{\frac{1}{n}}
Then I applied the Nth term test and used L'Hoptials rule so \lim_{n\to\infty} \frac{\cos(\frac{1}{n})\frac{-1}{n^2}}{\frac{-1}{n^2}}
The \frac{-1}{n^2} cancel out and the lim_{n\to\infty} \cos(\frac{1}{n}) is 1 which by the nth term test is divergent. Is that a legitimate proof of divergence?