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#1 2008-01-24 16:30:51

clooneyisagenius
Member
Registered: 2007-03-25
Posts: 56

Series and partial sum...

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:
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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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#2 2008-01-24 17:06:28

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Series and partial sum...

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 that’s

Therefore

That’s my conclusion.

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#3 2008-01-24 17:23:41

clooneyisagenius
Member
Registered: 2007-03-25
Posts: 56

Re: Series and partial sum...

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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#4 2008-01-24 17:28:39

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Series and partial sum...

So

is the answer? Thank you too! big_smile kiss

I wasn’t 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 wasn’t sure if I got it right. Thanks for confirming that the answer is correct. smile

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