My Math Forum infinite series#2

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 March 31st, 2011, 05:04 PM #1 Member   Joined: Jan 2010 Posts: 82 Thanks: 0 infinite series#2 I've been banging my head against the wall on this one, mainly because I feel like I'm very close. $\sum_{k=1}^{\infty}\frac{4}{4k^2+4k-3}$ I factored a 4 out of the bottom and then factored the denominator into two non-reduceable polynomials. aside#1 - $\frac{1}{(k+\frac{3}{2})(k-\frac{1}{2})}$ from there I factored out one-half from each poly and used partial fractions to cogitate: $1 \= \ A(2k-1) \ + \ B(2k+3)$ $let \ k \= \ \frac{-3}{2}, \ then A=\frac{-1}{4}, \ let \ k=\frac{1}{2}, \ then \ B=\frac{3}{4}$ $\therefore \ \sum_{k=1}^{5}\frac{4}{4k^2+4k-3} \ = \ \frac{1}{4} \sum_{k=1}^{5} \ [\frac{3}{2k-1} \ - \ \frac{1}{2k+3}] \ = \ \frac{1}{4}(3-\frac{1}{5}+1-\frac{1}{7}+\frac{3}{5}-\frac{1}{9}+\frac{3}{7}-\frac{1}{11}+\frac{3}{9}-\frac{1}{13})$ this means we have $1+\frac{1}{4}(\frac{2}{5}+\frac{2}{7}+\frac{2}{9}. ...\frac{2}{2n-1})$ does this mean that we can say $S_3 \= \ 1+ \frac{1}{4}(\frac{2}{2n-1})$ If so, how can we proceed? I'm not really sure if I have the tools to solve from here or perhaps I've just made a fundamental error that's leaving me exasperated, but I've thought about this quite a bit without much success. This is a take home quiz, so please don't solve this for me. thanks, Brad
 March 31st, 2011, 05:34 PM #2 Member   Joined: Jan 2010 Posts: 82 Thanks: 0 Re: infinite series#2 ugh. my partial fraction was wrong. it should have been $for \ k=\frac{1}{2}, \ B=\frac{1}{4}$ this solves most of my problem. b-
 March 31st, 2011, 05:53 PM #3 Member   Joined: Jan 2010 Posts: 82 Thanks: 0 Re: infinite series#2 it helps to spend 15 minutes away from something. It's solved. I'll post the answer in a bit for anyone else who's sadistic enough to be curious about my (fundamentally unsound) mistakes. b-
 March 31st, 2011, 06:46 PM #4 Member   Joined: Jan 2010 Posts: 82 Thanks: 0 Re: infinite series#2 $\sum_{k=1}^{\infty} \frac{4}{4k^2+4k-3} \rightarrow \sum \frac{1}{2k-1}-\frac{1}{2k+3}$ $\sum_{k=1}^{5}=1 -\frac{1}{5}+\frac{1}{3}-\frac{1}{7}+\frac{1}{5}-\frac{1}{9}+\frac{1}{7}-\frac{1}{11}+\frac{1}{9}-\frac{1}{13}$ notice....$a_k=4/3+\frac{-1}{2k+3}-\frac{1}{2(k+1)+3}$ $\lim_{k-=>\infty} \ \frac{4}{3}-\frac{1}{2n+3}-\frac{1}{2n+5} \= \ \frac{4}{3}$ $\therefore \ \sum_{k=1}^{\infty}\frac{4}{4k^2+4k-3} \ = \ \frac{4}{3}$

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