Comparison test for convergence problem: why is this incorrect?

[skyturnred]

Homework Statement

The original question is posted on my online-assignment. It asks the following:

Determine whether the following series converges or diverges:

\sum^{\infty}_{n=1}\frac{3^{n}}{3+7^{n}}

There are 3 entry fields for this question. One right next to the series above with the following options:

either \succ or \prec then next to that there is a field in which to input the thing that I am going to compare it to.

the third and final field I choose divergent or convergent

Homework Equations

The Attempt at a Solution

So I compared it to (\frac{4^{n}}{6^{n}}) because the original series is clearly less than this one. By doing the comparison test I determined that the series converges.

So I get "less than" right and "converges" right but I didn't get the other part right. Isn't it true that the original series is less than the one I decided above? Or was I wrong somewhere else?

 

[lanedance]

Everything you said sounds trues, though I'm not sure I understand the question?

Also, why wouldn't you just compare to \sum_n \frac{3^n}{7^n}?

 

[skyturnred]

That must be it.. I was just slight confused as to whether it was bigger or smaller than the original series, so I wanted to be EXTRA sure by choosing the one I mentioned above. I guess that must have been where I went wrong. But still, is what I did above correct? I realize that I could have made an "easier" decision for what to compare it to, but isn't what I chose still correct?

Thanks!

 

[lanedance]

yeah looks ok to me, generally want to choose the easiest to compare and the closest possible.

For the pupose of the convergence, as long as for some n>N each term yn> xn and (sum yn) converges, then (sum xn) converges

 

[kai_sikorski]

If it's an automatically graded online assignment the software might be hard-coded to only accept the most natural choice. What you did is correct though.