Does the given series converge absolutely or conditionally?

whatlifeforme
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Homework Statement


Determine either absolute convergence, conditional convergence or divergence for the series.

Homework Equations


\displaystyle \sum^{∞}_{n=1} \frac{(-1)^n}{5n^{1/4} + 5}

The Attempt at a Solution


It converges conditionally i know, but i can't figure out how.
1. I applied the alternating series test and concluded that the series converges.
2. I attempted to use the Ratio test for absolute convergence, but l'hopitals is just going in circles. I'm not getting anywhere; the limit n->∞ is ∞/∞.
 
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whatlifeforme said:

Homework Statement


Determine either absolute convergence, conditional convergence or divergence for the series.


Homework Equations


\displaystyle \sum^{∞}_{n=1} \frac{(-1)^n}{5n^(1/4) + 5}


The Attempt at a Solution


It converges conditionally i know, but i can't figure out how.
1. I applied the alternating series test and concluded that the series converges.
2. I attempted to use the Ratio test for absolute convergence, but l'hopitals is just going in circles. I'm not getting anywhere; the limit n->∞ is ∞/∞.

If you mean (-1)^n/(5n^(1/4)+5) try a comparison test with a p-series. The ratio test won't help you.
 
whatlifeforme said:

Homework Statement


Determine either absolute convergence, conditional convergence or divergence for the series.


Homework Equations


\displaystyle \sum^{∞}_{n=1} \frac{(-1)^n}{5n^(1/4) + 5}
Is this what you meant?
$$\sum^{∞}_{n=1} \frac{(-1)^n}{5n^{1/4} + 5} $$
whatlifeforme said:

The Attempt at a Solution


It converges conditionally i know, but i can't figure out how.
1. I applied the alternating series test and concluded that the series converges.
2. I attempted to use the Ratio test for absolute convergence, but l'hopitals is just going in circles. I'm not getting anywhere; the limit n->∞ is ∞/∞.
 
Mark44 said:
Is this what you meant?
$$\sum^{∞}_{n=1} \frac{(-1)^n}{5n^{1/4} + 5} $$

thanks. i fixed it.
 
Dick said:
If you mean (-1)^n/(5n^(1/4)+5) try a comparison test with a p-series. The ratio test won't help you.

how do i prove the p-series? use an integral test? i don't think my instructor will let us simply identify a p-series but wants us to rather prove it.
 
whatlifeforme said:
how do i prove the p-series? use an integral test? i don't think my instructor will let us simply identify a p-series but wants us to rather prove it.
A p-series (below) converges or diverges, depending on the value of p.
$$ \sum_{n = 1}^{\infty} \frac{1}{n^p}$$
Use the integral test, with f(x) = x-p, and look at cases for p < 1, p = 1, and p > 1. There might already be a proof of this in your book.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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