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Question: Prove that the partial sums are always greater than or equal to 1 once we have at least fiver terms. What number does this series appear to approach?
Series: 1 + (1/5) - (1/7) - (1/11) + (1/13) + (1/17) - ...
I found that:
But it is also given in the book that the sum is from k=1 to k with:
[(-1) ^ [Floor((k-1)/2)] ] / [6*(Floor(k/2)) + (-1)^(k-1)] (sorry not being in code - i'll work on it but wanted to get this up)
Any ideas on where to go?
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I do know that
(obtained by putting x=1 in the Maclaurin series of tan[sup]−1[/sup](x)).
So you just need to work out
Well thats
Therefore
Thats my conclusion.
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Wow. I feel like it was just one of those problems where I went right past something that wouldn't have been that difficult to making the problem very difficult. Thanks Jane. I knew the answer was pi/3 but you've helped and made it completely clear!!
I had a similar problem to do and thanks to you I've already done it! You are a great help!
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So
is the answer? Thank you too!I wasnt very sure about my working. I tried the sum of the first few terms on a calculator, but the convergence appears to be very slow so I wasnt sure if I got it right. Thanks for confirming that the answer is correct.
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