Determine whether or not the series is convergent or divergent. If it is convergent, find its sum.(adsbygoogle = window.adsbygoogle || []).push({});

1) 1/3 + 1/6 + 1/9 + 1/12 + 1/15 + ...

inf.

2) E (n-1)/(3n-1)

n=1

Okay, I'm having trouble with determining the convergence/divergence. My book states that, "If the series

inf.

E Asubn

n=1

is convergent, then lim as n goes to inf. asub n=0."

and it also states that, "If lim as n goes to inf. asubn does nto exist or if lim as n goes to inf. does not equal 0, then the series

inf.

E Asubn

n=1

is divergent.

----

Here's where I'm having trouble. I set up the sum series for the first question as:

inf.

E Asubn 1/(3n)

n=1

Then I took the limit as n goes to inf. of 1/(3n) and got 0. So by the book, wouldn't this make it convergent? However my book has the answer as being divergent.

Then for the second question, I took the limit as n goes to inf. of (n-1)/(3n-1) to be 1/3. So wouldn't this make it convergent as well? The book has the answer as being divergent.

If someone would articulate the convergent/divergent test a little better it would be greatly appreciated. Thanks!

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# Convergence/Divergence of a Series

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