Convergence/divergence of series

[Benny]

<br /> \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\frac{{n\pi }}{2}} \right)}}{{11 + 8n}}} <br />

I know that the numerator oscillates between -1 and 1 but there are some values of n for which the sine term also takes on the value of zero. So I can't find an explicit form for the numerator which means I can't use the alternating series test. I can't think of any other tests to use since the expression inside the summation takes on negative values 'regularly.'

<br /> \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^n \frac{1}{{n^{1 + \frac{1}{n}} }}} <br />

Hmm...this one is a bit trick so basically I just hoped that the alternating series test would yield something simple.

<br /> n^{1 + \frac{1}{n}} \ge n^{1 + \frac{1}{{n + 1}}} \Leftrightarrow \frac{1}{{n^{1 + \frac{1}{n}} }} \le \frac{1}{{n^{1 + \frac{1}{{n + 1}}} }}<br />

<br /> a_n \le a_{n + 1} <br />

The terms are not decreasing so the series diverges? My caculator suggests otherwise. Again, I'm not sure about this one.

Can someone help me out with these two series?

 

[HallsofIvy]

You can just ignore the 0 terms. If you want be precise, since sin(\frac{n\pi}{2}) is 0 whenever n is even, replace n with 2m+1. The series is exactly the same as
\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {\frac{{(2m+1)\pi }}{2}} \right)}}{{11 + 8(2m+1)}}}
That "skips" the 0 terms and is an alternating series with decreasing terms (in absolute value) and so converges.

 

[Hammie]

For the second series, finding the limit of the An may help.

This combined with the fact that the series is alternating may say something..

 

[Benny]

Thanks for the help guys. I'll try to finish them off now.