Sum of n*cos(1/n) over (2n+1) - Divergence Explained | Simplified Series

[frasifrasi]

Question asks to evaluate sum from 1 to infinity of n*cos(1/n) over (2n+1)

--> I am not sure how to do this.

I tried simplifying it to n/(2n+1) since the cos term is approaching 1 as n --> infinity. If i take the limit for that i get 1/2, so I concluded the series diverges.
--> but the bottom term is bigger than the top term, so why wouldn't it converge?

Thank you(it has been a while since I learned this so I am a bit confused).

 

[Dick]

Why would it converge just because the bottom term is larger than the top? You've already reached the correct conclusion. The limit of the nth term is 1/2. The sum can't possibly converge.

 

[frasifrasi]

I am saying, if i look at n/(2n+1), as n gets higher, the bottom is growing faster, so wouldn't it go to 0?

 

[Dick]

There is nothing wrong with posing theories like "If the denominator of a fraction is growing faster than the numerator then the limit is zero." We are doing mathematics here, so the next step is to test it. Take some sample terms like n=1000. 1000/2001. n=1000000. 1000000/2000001. They don't look like they are going to zero to me. It looks like they are approaching 1/2. So your 'theory' must be wrong. Try it with n/(2n).