1. ## Series Evaluation

I need to evaluate this for convergence/divergence:

$\sum_{n=1}^\infty \frac{n+5}{(n^7+n^2)^{1/3}}$

I tried a Limit Comparison Test with $\frac{1}{n^{7/3}}$ in the denominator, but that didn't get me anywhere. I don't see another function to compare this to.

Can anybody suggest what I might try?

Thanks.

2. ## Re: Series Evaluation

Originally Posted by joatmon
I need to evaluate this for convergence/divergence:

$\sum_{n=1}^\infty \frac{n+5}{(n^7+n^2)^{1/3}}$

I tried a Limit Comparison Test with $\frac{1}{n^{7/3}}$ in the denominator, but that didn't get me anywhere. I don't see another function to compare this to.

Can anybody suggest what I might try?

Thanks.
Maybe the integral test - the integral can be evaluated numerically...

3. ## Re: Series Evaluation

Oh boy. That's not going to be a fun one to evaluate. Thanks!

4. ## Re: Series Evaluation

Ever try some algebra?

$\frac{n+5}{(n^{7}+n^{2})^{\frac{1}{3}}} = \frac{n+5}{n\cdot(n^{4}+\frac{1}{n})^{\frac{1}{3}} }=\frac{1+\frac{5}{n}}{(n^{4}+\frac{1}{n})^{\frac{ 1}{3}}}<\frac{1+\frac{5}{n}}{n^{\frac{4}{3}}}$

It can get ugly, but it often leads somewhere useful.

Thanks!

6. ## Re: Series Evaluation

Hello, joatmon!

I need to evaluate this for convergence/divergence:

. . $\sum_{n=1}^{\infty} \frac{n+5}{(n^7+n^2)^{1/3}}$

For $n > 5$, we have: . $n+5 \:<\: 2n$ .[1]

We have: . $n^7+n^2 \:>\:n^7 \quad\Rightarrow\quad (n^7+n^2)^{\frac{1}{3}} \:>\:(n^7)^{\frac{1}{3}}$

. . Then: . $\frac{1}{(n^7 + n^2)^{\frac{1}{3}}} \:<\:\frac{1}{n^{\frac{7}{3}}}$ .[2]

Multiply [1] and [2]: . $(n+5)\cdot\frac{1}{(n^7 + n^2)^{\frac{1}{3}}} \;<\; 2n\cdot\frac{1}{n^{\frac{7}{3}}}$

. . . . . . . . We have: . . . . . . . $\frac{n+5}{(n^7+n^2)^{\frac{1}{3}}} \;<\;\frac{2}{n^{\frac{4}{3}}}$

Hence: . $\sum^{\infty}_{n=1}\frac{n+5}{(n^7+n^2)^{\frac{1}{ 3}}} \;<\;\sum^{\infty}_{n=1}\frac{2}{n^{\frac{4}{3}}}$

The given series is less than a convergent p-series.

. . Therefore, it converges.

7. ## Re: Series Evaluation

Well done. Thanks!

I was thinking about this a little more and I think that TKHunny's solution doesn't quite work because the final comparative function is not quite a p-series, so it doesn't definitively establish the original function as convergent. Then I saw Soroban's effort, and now I see how to get there.

Thanks to both of you!

8. ## Re: Series Evaluation

Originally Posted by joatmon
Well done. Thanks!

I was thinking about this a little more and I think that TKHunny's solution doesn't quite work because the final comparative function is not quite a p-series, so it doesn't definitively establish the original function as convergent. Then I saw Soroban's effort, and now I see how to get there.

Thanks to both of you!
No, you have come to the wrong conclusion. Mine doesn't quite work because I didn't do ALL the work for you. I left you some room to explore and learn on your own, through your personal effort.