Series

[canvas]

Determine the convergence/divergence of the following series.
Am I right?

[math]\sum_{n=1}^\infty\frac{1}{n}\cos(\frac{1}{n})\\\\\text{let's take}\,\,b_n=\frac{1}{n}\\\\\frac{a_n}{b_n}=\cos(\frac{1}{n})\longrightarrow1\\\\b_n\,\,\text{diverges, then}\,\,a_n\,\,\text{also diverges}[/math]

 

[Steven G]

I think that you failed to define a_n?

If n is sufficiently large, then what can you say about cos(1/n)

 

[pka]

Determine the convergence/divergence of the following series.
Am I right?

[math]\sum_{n=1}^\infty\frac{1}{n}\cos(\frac{1}{n})\\\\\text{let's take}\,\,b_n=\frac{1}{n}\\\\\frac{a_n}{b_n}=\cos(\frac{1}{n})\longrightarrow1\\\\b_n\,\,\text{diverges, then}\,\,a_n\,\,\text{also diverges}[/math]

Have you studied the limit comparison test?

 

[canvas]

Have you studied the limit comparison test?

how to do that ?

 

[Dr.Peterson]

Determine the convergence/divergence of the following series.
Am I right?

[math]\sum_{n=1}^\infty\frac{1}{n}\cos(\frac{1}{n})\\\\\text{let's take}\,\,b_n=\frac{1}{n}\\\\\frac{a_n}{b_n}=\cos(\frac{1}{n})\longrightarrow1\\\\b_n\,\,\text{diverges, then}\,\,a_n\,\,\text{also diverges}[/math]

how to do that ?

From your work, I was assuming you were using the limit comparison test!

What method are you using? And why do you have to ask about it when pka gave you a link?

 

[canvas]

I'm doing it by the limit comparison test, where is the mistake?

 

[Dr.Peterson]

I'm doing it by the limit comparison test, where is the mistake?

I don't think there's a mistake, but you need to say what you're doing. What you said is not enough, if a reader can't even be sure what test you are using!

As least, say what [imath]a_n[/imath] and [imath]b_n[/imath] mean, what role they play, and what theorem you are using to get what conclusion.

 

[canvas]

Limit comparison test:

[math]a_n=\frac{1}{n}\cos(\frac{1}{n})\\\\b_n=\frac{1}{n}\\\\c=\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{1}{n}\cos(\frac{1}{n})}{\frac{1}{n}}=\lim_{n\to\infty}\cos(\frac{1}{n})=1\in(0,+\infty)\\\\\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty\frac{1}{n}\,\,-\,\,\text{diverges, so also} \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{1}{n}\cos(\frac{1}{n})\,\,-\,\,\text{diverges}[/math]

 

[Dr.Peterson]

Limit comparison test:

[math]a_n=\frac{1}{n}\cos(\frac{1}{n})\\\\b_n=\frac{1}{n}\\\\c=\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{1}{n}\cos(\frac{1}{n})}{\frac{1}{n}}=\lim_{n\to\infty}\cos(\frac{1}{n})=1\in(0,+\infty)\\\\\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty\frac{1}{n}\,\,-\,\,\text{diverges, so also} \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{1}{n}\cos(\frac{1}{n})\,\,-\,\,\text{diverges}[/math]

Much better. Now we can say with confidence that you are right.

 

[canvas]

Thank you so much, have a good night

 

[lookagain]

Much better. Now we can say with confidence that you are right.

Hold on, please. WolframAlpha has graphics and a limit number. It states that this
series converges.

 

[Dr.Peterson]

Hold on, please. WolframAlpha has graphics and a limit number. It states that this
series converges.

That's odd. It says 5.31884, which turns out to be the sum of just the first 200 terms. When I take it to 2000 terms, I get
7.619169 (and counting). And WA agrees on both of those.

I wonder what WA is thinking? It doesn't appear to know convergence tests.

 

[blamocur]

Another way to look at the problem (direct comparison test):
[math]a_n = \frac{1}{n} \cos\left(\frac{1}{n}\right) \geq \frac{1}{n}\cos 1 \Longrightarrow \sum a_n \geq \cos 1 \sum \frac{1}{n}[/math]