# Is there an easier way?

1. Sep 16, 2006

### G01

Is the following limit, conditionally covergent, abolutely convergent, or divergent?

$$\sum_{n=1}^{\infty} \frac{(-1)^n e^{1/n}}{n^3}$$

To solve this i used the ratio test with help from L'Hospitals Rule. That took me forever and was really complicated. Can anyone think of an easier way?

2. Sep 16, 2006

### StatusX

Say you want to show that it's absolutely convergent. Then you need to show the series formed by taking the absolute value of each term of this series is convergent. The terms of that series approach 1/n3 from above. Can you show that:

$$\sum_{n=1}^\infty \frac{1}{n^3}$$

converges? Then you have one more step left.

3. Sep 16, 2006

### arildno

It is an alternating series with terms decreasing in absolute value towards zero.
That is sufficient to tell you whether the series diverges or converges.

4. Sep 16, 2006

### G01

Hmm, can I use a comparison test between

$$\sum_{n=1}^{\infty} \frac{(-1)^n e^{1/n}}{n^3}$$
and
$$\sum_{n=1}^\infty \frac{1}{n^3}$$
?

5. Sep 16, 2006

### arildno

Letting the n'th term be called a(n), then you have:
$$|a(n)|\leq\frac{e}{n^{3}}$$
what does this tell you?