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So I have the infinite sum: $$\sum_{n=1}^\infty (-1)^{n-1} nx^{2n}$$ I have found in a previous part that the convergence interval is $(-1,1)$. I know that the sum should equal $$\frac {x^2}{{(x^2 +1)}^2}$$ when $|x|\lt1$ But I'm not sure how I can show this from the sum. |
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Recall the alternating geometric series: $$\frac1{1+x}=1-\sum_{n=1}^\infty(-1)^{n-1}x^n$$ Differentiate both sides and multiply both sides by $-x:$ $$\frac x{(1+x)^2}=\sum_{n=1}^\infty(-1)^{n-1}nx^n$$ Let $x\mapsto x^2$ and we finally get $$\frac{x^2}{(1+x^2)^2}=\sum_{n=1}^\infty(-1)^{n-1}nx^{2n}$$ |
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I thought it might be instructive to present an approach that uses pre-calculus tools only. To that end, we proceed. Note that $n=\sum_{m=1}^{n-1} (1)$. Therefore, we have $$\begin{align} \sum_{n=1}^\infty (-1)^{n-1} nx^{2n}&=-\sum_{n=1}^\infty\sum_{m=1}^n(1) (-x^2)^n\\\\ &=-\sum_{m=1}^\infty\sum_{n=m}^\infty (-x^2)^n \,\,\,\,\dots\,\,\text{interchange summation order}\\\\ &=-\sum_{m=1}^\infty\frac{(-x^2)^m}{1+x^2}\,\,\,\,\,\,\,\,\,\,\dots\,\,\,\text{Sum inner geometric series}\\\\ &=\frac{x^2}{(1+x^2)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dots\,\,\text{Sum geometric series} \end{align}$$ And we are done! |
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\begin{align} \sum_{n=1}^\infty (-1)^{n-1} nx^{2n} = x^2 - {}2x^4 + 3x^6 - 4x^8 + 5x^{10} - \cdots \end{align} $$ \begin{array}{cccccccccc} n=1 & & n=2 & & n=3 & & n=4 & & n=5 & & \cdots \\[10pt] \hline x^2 & + & (-x^4) & + & x^6 & + & (-x^8) & + & x^{10} & + & \cdots \\[10pt] & & (-x^4) & + & x^6 & + & (-x^8) & + & x^{10} & + & \cdots \\[10pt] & & & & x^6 & + & (-x^8) & + & x^{10} & + & \cdots \\[10pt] & & & & & & (-x^8) & + & x^{10} & + & \cdots \\[10pt] & & & & & & & & x^{10} & + & \cdots \\[10pt] & & & & & & & & & & \cdots \end{array} $$ First find the sum of each row separately. That's easy because each one is a geometric series. Then find the sum of all those resulting sums. That also easy because that is also a geometric series. |
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Rewrite $n$ as binomial coefficient, then "negate the upper index" $\binom nk=(-1)^k\binom{k-n-1}k$ for $k=n-1$: $$ \sum_{n=1}^\infty (-1)^{n-1} nx^{2n} =\sum_{n=1}^\infty (-1)^{n-1} \binom n{n-1}(x^2)^n =\sum_{n=1}^\infty \binom {-2}{n-1}(x^2)^n\\ =\sum_{m=0}^\infty \binom {-2}m(x^2)^{m+1} =x^2(1+x^2)^{-2}, $$ the final step using Newton's (generalised) binomial formula for exponent $-2$. |
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