Question

Can someone help me set up this series to determine convergence/divergence

Answer 1

\[\sum_{n=1}^{\infty} (4+3^n)/2^n\]

I know that I need to use comparison test and then use p-series.. Or could I just use geometric series and say that since (4+3^n)/2^n < 3^n/2^n and since (3/2)^n and 3/2>1 then the series is divergent.. I think I just figured it out on my own, but can anyone clarify this?

Answer 2

Direct comparison.
The series:
\[\sum_{n=1}^{\infty} \left(\frac{3}{2} \right)^n < \sum_{n=1}^{\infty} \frac{(4+3^n)}{2^n}\]
Rewrite the first one as:
\[\frac{3}{2} \sum_{n=1}^{\infty} \left( \frac{3}{2} \right)^{n-1}\]
Since 3/2>1 the series diverges by the geometric series test.
Since the series is smaller than the one of study, since a smaller series diverges the larger series must diverge.

Answer 3

Yea... As I was typing the answer appeared to me lol figured I'd ask anyway to be sure I was on the right track. Thanks

Answer 4

Direct comparison.
The series:
\[\sum_{n=1}^{\infty} \left(\frac{3}{2} \right)^n < \sum_{n=1}^{\infty} \frac{(4+3^n)}{2^n}\]
Rewrite the first one as:
\[\frac{3}{2} \sum_{n=1}^{\infty} \left( \frac{3}{2} \right)^{n-1}\]
Since 3/2>1 the series diverges by the geometric series test.
Since the series is smaller than the one of study, since a smaller series diverges the larger series must diverge.

Answer 5

No problem :P