Question on a P-series problem

[Edder]

This problem states: for what values of P is the series convergent?

\(\displaystyle \sum_{n=1}^{\infty}\left( \frac{4n^6}{3n^8+8}\right)^{p}\)

How would I go about solving this? I obviously can't used the root test, so any ideas?


Any feedback appreciated, thanks.

 

[daon2]

If p>=0,

\(\displaystyle \left(\frac{4n^6}{3n^8+8n^8}\right)^p \le \left(\frac{4n^6}{3n^8+8}\right)^p \le \left(\frac{4n^6}{3n^8}\right)^p\)

 

[Deleted member 4993]

This problem states: for what values of P is the series convergent?

\(\displaystyle \sum_{n=1}^{\infty}\left( \frac{4n^6}{3n^8+8}\right)^{p}\)

How would I go about solving this? I obviously can't used the root test, so any ideas?


Any feedback appreciated, thanks.

\(\displaystyle \sum_{n=1}^{\infty}\left( \frac{4n^6}{3n^8+8}\right)^{p} \le \sum_{n=1}^{\infty}\left( \frac{4}{3}\frac{1}{n^2}\right)^{p}\)

Now tell us for which value of 'p' - the reduced series is convergent?

 

[Edder]

\(\displaystyle \sum_{n=1}^{\infty}\left( \frac{4n^6}{3n^8+8}\right)^{p} \le \sum_{n=1}^{\infty}\left( \frac{4}{3}\frac{1}{n^2}\right)^{p}\)

Now tell us for which value of 'p' - the reduced series is convergent?

By the rule of the p-series, if p>1 then the series is convergent. So by looking at reduced series \(\displaystyle \frac{1}{n^2}\), the only value that will make the p=1 is 1/2. So I am assuming the series is convergent for the values of p>1/2. Is this correct?