# Thread: infinite series

1. ## infinite series

so if I use the comparison test to determine whether the infinite series, $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{2n^3 + 7n}}$, and using $\displaystyle \frac{1}{n^{\frac{3}{2}}}$ as the comparing series....should the initial infinite series be convergent or divergent

my belief is that it should be divergent because $\displaystyle \frac{1}{n^{\frac{3}{2}}}$ is divergent and it is greater than $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{2n^3 + 7n}}$

Thanks in advance for any help provided

2. ## Response

This is what I found

The p-series is given by
Σ[1/(n^p)] = 1/(1^p) + 1/(2^p) + 1/(3^p) + ...
where p > 0 by definition.
If p > 1, then the series converges.
If 0 < p <= 1 then the series diverges.

Since 3/2 > 1, the series should converge. Hope this helps.

3. Originally Posted by ajj86
This is what I found

The p-series is given by
1/(n^p) = 1/(1^p) + 1/(2^p) + 1/(3^p) + ...
where p > 0 by definition.
If p > 1, then the series converges.
If 0 < p <= 1 then the series diverges.

Since 3/2 > 1, the series should converge. Hope this helps.

AH~ the p-series. good thinking 99.
So in effect the initial series converges as well....correct?

4. ## Response

You got it. Basically, your statement is correct if you substitute convergent for divergent in the sentence you wrote above:
my belief is that it should be convergent because $\displaystyle \frac{1}{n^{\frac{3}{2}}}$
is convergent and it is greater than

$\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{2n^3 + 7n}}$