Series convergent or divergent by Comparison Test.

**oceanmd** · March 13th 2009, 07:55 PM

Please deternine whether the series is convergent or divergent by Comparison Test.
Series sign (2^(n+1))/(3^(n+1))
My resoning: bn=2^n/3^n= (2/3)^n, r=2/3 less than 1, geometric series converges. It is true that an is less than bn, it means that the initial series converges.

If this is wrong, pleaseexplain how to do it by Comparison Test.
Thank you very much

**matheagle** · March 13th 2009, 08:15 PM

There's no need to compare this to anything.
This is a geometric with r=2/3. This is just $\biggl({2\over 3}\biggr)^{n+1}$ .
You can do the ratio test and get 2/3, but that's basically a limit comparison test to the geo's.
You can find what the sum converges to, but you need to tell me where n starts.

**oceanmd** · March 14th 2009, 08:21 AM

I do not have a choice in the method. The assignment says I must do it using Comparison Test. n=1,
Is the answer that the series converges correct?
Thank you

**matheagle** · March 14th 2009, 01:03 PM

Sure compare it to itself...

$\biggl({2\over 3}\biggr)^{n+1}= \biggl({2\over 3}\biggr)\biggl({2\over 3}\biggr)^n$ .

So

$\sum_{n=1}^{\infty}\biggl({2\over 3}\biggr)^{n+1}= \biggl({2\over 3}\biggr)\sum_{n=1}^{\infty}\biggl({2\over 3}\biggr)^n<\sum_{n=1}^{\infty}\biggl({2\over 3}\biggr)^n<\infty$ .

But the answer is...

$\biggl({2\over 3}\biggr)\biggl({2/3\over 1-2/3}\biggr)={4/9\over 1/3}={4\over 3}$ .

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