Determine convergence of a series with sine

[Mampac]

Hello,

I know this may seem a very easy problem that is solved in couple of steps, but I have a question... The TA who solved this during a problem solving session focused on the function inside the sine. Since 2 times an integral of 1/n is divergent (power of n is <= 1), then the whole function is divergent.

However, a thought crossed my mind about the limit of the fraction -- isn't it 0 as n tends to infinity? 2 over infinity + 3 must yield 0 and sin(0) is 0 thus the sum is 0. Doesn't that mean the series are finite thus convergent?

 

[ISTER_REG]

I would like to interject here that perhaps one idea is to use the knowledge to harmonic series. We know for example that [MATH]\sum_{n=1}^{\infty} \frac{1}{n}[/MATH] diverges (you mentioned that). Furthermore the sum [MATH]\sum_{n=1}^{\infty} \frac{1}{n+1}[/MATH] also diverges by the limit comparison test. The last one looks more like the fraction in the sine and maybe its an idea to argue with that knowledge, that this is also divergent... Proof me wrong

 

[lex]

2 over infinity + 3 must yield 0 and sin(0) is 0 thus the sum is 0. Doesn't that mean that the series are finite thus convergent?

The individual terms tending to 0 does not mean that the sum of the series is 0, nor that the series converges.
E.g. for [MATH]\sum_{n=1}^{\infty} \frac{1}{n}[/MATH], the individual terms [MATH]\frac{1}{n}[/MATH] tend to 0, but the series diverges.

1. The present example can be shown to diverge by comparison with [MATH]\frac{1}{n}[/MATH].
It is fairly straightforward to show that: [MATH]\quad \text{ sin}\left(\frac{2}{x+3}\right)>\frac{1}{x}[/MATH] when [MATH]x>4[/MATH]You can then argue that the comparison test applies, and therefore [MATH]\sum_{n=1}^{\infty} \text{sin}\left(\frac{2}{n+3}\right)[/MATH] diverges.

OR
2. You can do the limit comparison test with [MATH]\frac{1}{n}[/MATH]Look at [MATH]\lim_{x\rightarrow\infty}{\frac{\sin\left(\frac{2}{x+3}\right)}{\frac{1}{x}}\ }[/MATH]