Root Test to an Infinite Series

**Beevo** · September 15th 2012, 03:45 PM

$\sum_{n=1}^{\infty} \frac{3n + 2}{n} ({8/5})^n$

The question asks me to compute the value of p for the series with the root test. I took the root test and got:

$\sum_{n=1}^{\infty} \frac{3n +2}{n} ({8/5})$ , as with the limit of n going to infinity. By that notion, I got 3 x (8/5), which is 24/5, but the answer is incorrect. Can anyone tell what I am doing wrong?

Thanks

**GJA** · September 15th 2012, 05:53 PM

Hi, Beevo.

When you say "the value of p" do you mean the numeric value the root test will give you?

When you do the root test you should take the nth root of the $a_{n}$ th term of the series. If you do this you get

$\sqrt[n]{\frac{3n+2}{n}\left(\frac{8}{5} \right)^{n}}=\sqrt[n]{\frac{3n+2}{n}}(8/5)$

To finish the problem we need to determine what the limit on the right side is.

Does this help? If you're still stuck let me know.

Good luck!

**Beevo** · September 15th 2012, 06:06 PM

Originally Posted by GJA

Hi, Beevo.

When you say "the value of p" do you mean the numeric value the root test will give you?

When you do the root test you should take the nth root of the $a_{n}$ th term of the series. If you do this you get

$\sqrt[n]{\frac{3n+2}{n}\left(\frac{8}{5} \right)^{n}}=\sqrt[n]{\frac{3n+2}{n}}(8/5)$

To finish the problem we need to determine what the limit on the right side is.

Does this help? If you're still stuck let me know.

Good luck!

No I eventually figured it out. I forgot to take the nth root of both factors.

Thanks for your input though, appreciate it.

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