Infinite Series
- Thread starter sickplaya
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- Apr 12, 2005
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Just add it all up.sickplaya said:
5x^2 + 9x^4 + 13x^6 + 17x^8 +..... = S
5x^4 + 9x^6 + 13x^8 + 17x^10 +..... = S*x^2
Subtracting
5x^2 + 4x^4 + 4x^6 + 4x^8 +..... = S*(1-x^2)
Judicious Factoring
5x^2 + (4x^4)*(1 + x^2 + x^4 +.....) = S*(1-x^2)
Can you add up the one in the parentheses and solve for S?
It's not pretty, but it works.
This may be little belated but here's a method:
You have \(\displaystyle \sum_{n=1}^{\infty}\(4n+1)x^{2n}\), x<1
=\(\displaystyle 4\sum_{n=1}^{\infty}nx^{2n}+\sum_{n=1}^{\infty}x^{2n}\)
It can be shown that:
\(\displaystyle \sum_{n=1}^{\infty}nx^{2n}=\sum_{n=0}^{\infty}x^{2n}\sum_{n=1}^{\infty}x^{2n}\)
\(\displaystyle \sum_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^{2}}=\frac{1}{1-(.85634884)^{2}}=3.75=\frac{15}{4}\)
Therefore, \(\displaystyle \sum_{n=1}^{\infty}x^{2n}=2.75=\frac{11}{4}\)
\(\displaystyle (4)(3.75)(2.75)=(4)(\frac{15}{4})(\frac{11}{4})=41.25\)
So we have: \(\displaystyle 41.25+2.75=44\)
You have \(\displaystyle \sum_{n=1}^{\infty}\(4n+1)x^{2n}\), x<1
=\(\displaystyle 4\sum_{n=1}^{\infty}nx^{2n}+\sum_{n=1}^{\infty}x^{2n}\)
It can be shown that:
\(\displaystyle \sum_{n=1}^{\infty}nx^{2n}=\sum_{n=0}^{\infty}x^{2n}\sum_{n=1}^{\infty}x^{2n}\)
\(\displaystyle \sum_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^{2}}=\frac{1}{1-(.85634884)^{2}}=3.75=\frac{15}{4}\)
Therefore, \(\displaystyle \sum_{n=1}^{\infty}x^{2n}=2.75=\frac{11}{4}\)
\(\displaystyle (4)(3.75)(2.75)=(4)(\frac{15}{4})(\frac{11}{4})=41.25\)
So we have: \(\displaystyle 41.25+2.75=44\)