Determine if the series converges or diverges: $$\sum_{n=1}^\infty \frac{\ln(n)}{n \left|\sin(n)\right|} = $$
We know that
$$ \sum_{n=1}^\infty \frac{1}i = \text{Divergent} $$
$$ \sum_{n=1}^1 \ln(i )= 0 $$
How to solve this?
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Determine if the series converges or diverges: $$\sum_{n=1}^\infty \frac{\ln(n)}{n \left|\sin(n)\right|} = $$ We know that $$ \sum_{n=1}^\infty \frac{1}i = \text{Divergent} $$ $$ \sum_{n=1}^1 \ln(i )= 0 $$ How to solve this? |
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Another way: also note $\frac{\log n }{n|\sin n|} \geq \frac{\log n}{n}$. Now look at $\sum_{n=1}^{\infty}\frac{\log n }{n}$. By intergral test if the integral diverges, then the sum diverges too. $\int_{1}^{\infty} \frac{\log x dx}{x} = \left. \frac{ \log^{2} x}{2} |^{\infty}_{1}\right.=\infty$. The integral diverges, hence the sum diverges too. Since the original sum is lower-bounded by this one, by comparison test it diverges too. |
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Hint: Since $|\sin n|\le 1$, and $\ln n \gt 1$ for $n\ge 3$, the $n$-th term of our sequence is greater than $\frac{1}{n}$ if $n\ge 3$. Now use the Comparison Test. |
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