Is there any closed form solution of the following series (assuming $0 \leq x \leq 1$)? $$2x^2 + 4x^4 + 6x^6 + 8x^8 \ldots$$
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Yes. Do you know any methods you can use to go from one power series to another? – Brian Moehring Jan 30 '21 at 23:20
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I know what $\sum_{n =0}^{\infty}nx^n$ should be. But can't figure out the series when $n$ takes on only even values. – Resting Platypus Jan 30 '21 at 23:22
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2All right. If you have $f(x) = \sum_{n=0}^\infty nx^n$, what is $f(x) + f(-x)$? – Brian Moehring Jan 30 '21 at 23:25
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We get only the even terms i.e. $2x^2+4x^4 \ldots $. Okay, so the only thing left to prove is that $f(x)$ is even? – Resting Platypus Jan 30 '21 at 23:59
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Hint: put $y = x^2$, then $$ 2x^2 + 4x^4 + 6x^6 + 8x^8 + \ldots = 2y(1 + 2y + 3y^2 + 4y^3 + \ldots) $$ You can get a closed formula for $s(y) = 1 + 2y + 3y^2 + 4y^3 + \ldots$ using the formula for the sum of a geometric series either by considering it as the derivative of the geometric series $y + y^2 + y^3 + y^4 + \ldots$ or by expanding $s(y) - ys(y)$.
Rob Arthan
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Thanks very much for an elegant solution. I used your hint relating to derivatives! – Resting Platypus Jan 31 '21 at 02:07
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some hints:
geometric-like series
substitute $w=x^2$
$\sum_{n=1}^\infty 2n x^{2n} = 2\sum_{n=1}^\infty n w^n = \ldots$
GhostAmarth
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$$\sum\limits_{n=1}^\infty2nx^{2n}=\sum\limits_{n=1}^\infty2\sum\limits_{m=n}^\infty x^{2n}=\sum\limits_{n=1}^\infty2\left(\frac{x^{2n}}{1-x^2}\right)=\frac2{1-x^2}\sum\limits_{n=1}^\infty x^{2n}=\frac2{1-x^2}\cdot\frac{x^2}{1-x^2}$$$$=\frac{2x^2}{(1-x^2)^2}$$
Rushabh Mehta
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