Convergence of sin(1/n) - sin(1/(n+1)) and arctg(n)

**NZAU1984** · October 12th 2012, 02:09 PM

Hi,

Still me with some limit problems...

Now I just don't any clues about what to do with trigonometric functions...

I have this :

$\sum_{n=1}^{\infty}sin(\frac{1}{n}) - sin(\frac{1}{n+1})$

I know it becomes this :

$\sum_{n=1}^{\infty}sin(\frac{1}{n}) - \sum_{n=1}^{\infty}sin(\frac{1}{n+1})$

I know that the first one will tend towards 0 since sin(1/infinity) = sin(0) = 0

The second one, believe it or not, I'm not so sure about how to find its limit.

But I know that the limits of those two blocks isn't the answer of the question, that I have to look at their "s_n"'s, but how would I be able to estimate sin(1) + sin(1/2) + sin(1/2) + ... and the other one that I won't write here ?

Same thing for the following, I'm clueless...

$\sum_{n=1}^{\infty}arctg(n)$

Please help, I'm kind of lost right now.

**Plato** · October 12th 2012, 02:59 PM

Originally Posted by NZAU1984

I have this :
$\sum_{n=1}^{\infty}sin(\frac{1}{n}) - sin(\frac{1}{n+1})$

$\sum_{n=1}^{\infty}arctg(n)$

Like the other you posted this is a collapsing sum.
If $S_N=\sum_{n=1}^{N}sin(\frac{1}{n}) - sin(\frac{1}{n+1})$ then $S_N = \left[ {\sin (1) - \sin \left( {\frac{1}{{n + 1}}} \right)} \right]$ .

So $\left( {S_N } \right) \to ?$

Because $(\arctan(n))\to\frac{\pi}{2}$ can $\sum_{n=1}^{\infty}\arctan(n)$ converge?

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