# Convergent or Divergent

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?

-

1
you need to remove 1 from right hand side of first eq. The other two can be reworded too. – Maesumi Feb 2 at 3:26

## 2 Answers

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.

-

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.

-

Got it thanks! That seems easier now – user1730308 Feb 2 at 3:18
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8597381114959717, "perplexity_flag": "middle"}