Hi, I was wondering if someone can help me with this problem.

Determine whether the series is convergent or divergent

\sum_{n=1}^ (infinity) 4 + (3^n) / (2^n)

Thank you to anyone who can help.

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- March 4th 2007, 12:10 PMWintaker99convergent or divergent?
Hi, I was wondering if someone can help me with this problem.

Determine whether the series is convergent or divergent

\sum_{n=1}^ (infinity) 4 + (3^n) / (2^n)

Thank you to anyone who can help. - March 4th 2007, 12:59 PMCaptainBlack
Divergent:

\sum (n=1 to infinity) [4 + 3^n] / (2^n) = \sum (n=1 to infinity) 4/ (2^n) + \sum (n=1 to infinity) 3^n / 2^n

................= \sum (n=1 to infinity) 4/ (2^n) + \sum (n=1 to infinity) (3/2)^n

The first of the sums in the above converges as its a geometric series with factor <1, while the second sum diverges as again it is a geometric series with factor >1.

RonL