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+--- Thread: Summing up Series (/showthread.php?tid=63235)
Summing up Series - Imagine - 2013-03-17
Given the series:
1/((n+1)(3^(n+1))
Prove that it converges or diverges. If it converges, find its sum.
I can prove that it converges, but how would I find the sum?
Summing up Series - Locked - 2013-03-17
You missed the most crucial information.
![[Image: crlxvwc.png]](http://mathurl.com/crlxvwc.png)
Summing up Series - Imagine - 2013-03-17
Ah, from 1 to infinity.
Summing up Series - Locked - 2013-03-17
Well, just find the partial sums and then take the limit of it as it approaches infinity.
Summing up Series - Imagine - 2013-03-17
Figured it out. It wasn't partial sums.
-ln(1 - x) = sum from 0 to infinity of (x^(n+1))/(n + 1)
In my case, x = 1/3.
So, the sum is -ln(2/3).
Summing up Series - Locked - 2013-03-17
It's -1/3 + ln(3/2)
Summing up Series - Imagine - 2013-03-17
My work:
![[Image: hkfwhCg.jpg]](http://i.imgur.com/hkfwhCg.jpg)
What am I missing?
EDIT: It starts at 0, not 1 once I looked over the problem again.
Summing up Series - Locked - 2013-03-17
Dat handwriting.
Anyway
If it's zero then it's ln(3/2) but:
-ln(2/3) = -ln(2) - (-ln(3)) = ln(3) - ln(2) = ln(3/2)
So it's the same thing.
Summing up Series - Stereo - 2013-03-28
Locked Wrote:Dat handwriting.
Anyway
If it's zero then it's ln(3/2) but:
-ln(2/3) = -ln(2) - (-ln(3)) = ln(3) - ln(2) = ln(3/2)
So it's the same thing.
Late to the party but (-1)*ln(2/3) = ln((2/3)^(-1)) = ln(3/2) is another way to show that.