# Math Help - Series Sum Question

1. ## Series Sum Question

$\sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4n\sqrt{n}+4}$

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

The problem is to decide if the series converge or diverge. I have tried the n-th term and integral tests but to no avail.

For the second problem, I have shown that the n-th term test does not diverge, but that does not nececarily show that it DOES converge, I believe.

2. On the first one it helps to realize $n\sqrt{n}=\sqrt{n^3}$.

3. ok, so...
$
\sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4\sqrt{n^3}+4}
$

Can this help to integrate? I don't see how to use the hint, sorry! Maybe I am forgetting a rule for radicals that I could use, I will take another look. Thanks though!

4. Originally Posted by matt.qmar
ok, so...
$
\sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4\sqrt{n^3}+4}
$

Can this help to integrate? I don't see how to use the hint, sorry! Maybe I am forgetting a rule for radicals that I could use, I will take another look. Thanks though!
it allows one to "see" that the nth term approaches $\frac{\sqrt{2}}{4}$ as $n \to \infty$

5. Originally Posted by matt.qmar
$\sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4n\sqrt{n}+4}$

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

The problem is to decide if the series converge or diverge. I have tried the n-th term and integral tests but to no avail.

For the second problem, I have shown that the n-th term test does not diverge, but that does not nececarily show that it DOES converge, I believe.

The general term of the first series doesn't even converge to zero so the series is divergent
]
For the second one you ahve 1/[n^(3/2)*e^n] <= 1/n^(3/2) and apply the comparison test.

Tonio

6. Thank you! I forgot the simple radical division rule. Much appriciated!