Help with series absolutely convergent...

Question

anonymous · Answer

For which values ​​of p is the series
$$\sum_{n=1}^{\infty}(-1)^n \frac{ n^p }{ n+1 }$$
absolutely convergent?

I should use the "Limit comparison test"

Please help!

goformit100 · Answer

Its easy remove summation and put integration sign and solve it.

anonymous · Answer

Is that the Limit comparison test!?!? @goformit100

experimentx · Answer

p<0

experimentx · Answer

compare it with $ \sum1/n^p $

anonymous · Answer

Like this @experiment : $$\frac{ \frac{ 1 }{ n^p } }{ \frac{ n^p }{ n+1 }}$$

anonymous · Answer

No, compare it with $$\sum_{n=1}^{\infty}\frac{ 1 }{ n^{1-p} }$$

anonymous · Answer

With the Direct Comparison Test?

anonymous · Answer

No, Limit Comparison Test

anonymous · Answer

Sorry drawar. But I dont understand. The Limit comparison test is: http://www.mathwords.com/l/limit_comparison_test.htm Agree? Can you help me a take it a little more step by step?

anonymous · Answer

Set the ratio, take the limit

experimentx · Answer

$$ \frac{n^p}{n+1} < \frac{n^p}{n} = \frac{1}{n^{1-p}}$$
now for what value of p does this series converge?

anonymous · Answer

p < 0 ..

experimentx · Answer

yes that's correct

anonymous · Answer

2 questions in the assignment are:

For which values ​​of p the series is convergent?

Is not it the same value?
Because if it is absolutely convergent then it is also convergent.

experimentx · Answer

no ... alternating you use Lebnitz test for alternating series. or p<1, the series converges conditionally. http://www.wolframalpha.com/input/?i=Sum%5B%28-1%29%5En%2Fn%5E%280.0005%29%2C+%7Bn%2C+1%2C+Infinity%7D%5D