Alternating Sums and Areas

Let $|x| < 1$ and $(0 < a < 1)$ .

$\sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}$

$(\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2}.$

$\sum_{n = 1}^{\infty} n^2 x^n = \sum_{j = 1}^{\infty} (\sum_{n = j}^{\infty} n x^n) =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} (j x^j) + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^j =$

$(\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x}) =$ $\dfrac{x^2 + x}{(1 - x)^3}$ .

$\sum_{n = 1}^{\infty} n^3 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^2 x^{n} =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{2x}{(1 - x)^2}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^2 + x}{(1 - x)^3}\sum_{j = 1}^{\infty} x^j = (\dfrac{1}{1 - x})(\dfrac{x^2 + x}{(1 - x)^3}) +$ $(\dfrac{2x}{(1 - x)^2})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^2 + x}{(1 - x)^3})(\dfrac{x}{1 - x}) =$ $\dfrac{x^3 + 4x^2 + x}{(1 - x)^4}$ .

$\sum_{n = 1}^{\infty} n^4 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^3 x^{n} =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{3x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{3(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}\sum_{j = 1}^{\infty} x^j =$

$= (\dfrac{1}{1 - x})( \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}) +$ $(\dfrac{3x}{(1 - x)^2})(\dfrac{x^2 + x}{(1 - x)^3}) + (\dfrac{3(x^2 + x)}{(1 - x)^3})(\dfrac{x}{(1 - x)^2}) +$ $\dfrac{x^3 + 4x^2 + x}{(1 - x)^4}(\dfrac{x}{1 - x}) = \dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}$ .

$\sum_{n = 1}^{\infty} n^5 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^4 x^{n} =$

$\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^4 x^j + \dfrac{4x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{6(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j^2 x^j + 4\dfrac{x^3 + 4x^2 + x}{(1 - x)^4}\sum_{j = 1}^{\infty} j x^j$

$+ \dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}\sum_{j = 1}^{\infty} x^j =$

$\dfrac{1}{1 - x}(\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5})$ + $\dfrac{4x}{(1 - x)^2}(\dfrac{x^3 + 4x^2 + x)}{(1 - x)^4}$ + $\dfrac{6(x^2 + x)}{(1 - x)^3}(\dfrac{x^2 + x}{(1 - x)^3}) + \dfrac{4(x^3 + 4x^2 + x)}{(1 - x)^4}(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5})(\dfrac{x}{1 - x})$

$= \dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6}$

$\implies \boxed{g(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} n^5 x^n = - \sum_{n = 1}^{\infty} n^5 (-x)^n = }$

$\boxed{-(\dfrac{-x^5 + 26x^4 - 66x^3 + 26x^2 - x}{(1 + x)^6}) = \dfrac{x^5 - 26x^4 + 66x^3 - 26x^2 + x}{(1 + x)^6}}$

$\sum_{n = 1}^{\infty} n^6 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^5 x^{n} =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^5 x^j + \dfrac{5x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^4 x^j + \dfrac{10(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{10(x^3 + 4x^2 + x)}{(1 - x)^4}\sum_{j = 1}^{\infty} j^2 x^j$

$+ \dfrac{5(x^4 + 11x^3 + 11x^2 + x)}{(1 - x)^5}\sum_{j = 1}^{\infty} j x^j +$ $(\dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6})\sum_{j = 1}^{\infty} x^j =$

$\dfrac{1}{1 - x}(\dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6}) +$ $\dfrac{5x}{(1 - x)^2}(\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}) + \dfrac{10(x^2 + x)}{(1 - x)^3}(\dfrac{x^3 + 4x^2 + x}{(1 - x)^4})$

$+ \dfrac{10(x^3 + 4x^2 + x)}{(1 - x)^4}(\dfrac{x^2 + x}{(1 - x)^3}) + \dfrac{5(x^4 + 11x^3 + 11x^2 + x)}{(1 - x)^5}(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6})(\dfrac{x}{1 - x})$ .

$= \dfrac{x^6 + 57x^5 + 302x^4 + 302x^3 + 57x^2 + x}{(1 - x)^7}$

$\implies \boxed{f(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} n^5 x^n = - \sum_{n = 1}^{\infty} n^6 (-x)^n = }$

$\boxed{-(\dfrac{x^6 - 57x^5 + 302x^4 - 302x^3 + 57x^2 - x}{(1 + x)^7}) = \dfrac{-x^6 + 57x^5 - 302x^4 + 302x^3 - 57x^2 + x}{(1 + x)^7}}$

Let $u = 1 + x \implies du = dx \implies$

$\int_{a - 1}^{0} g(x) dx = -\int_{1}^{a} \dfrac{u^5 - 31u^4 + 180u^3 - 390u^2 + 360u - 120}{u^6} du =$

$-\int_{1}^{a} \dfrac{1}{u} - 31u^{-2} + 180u^{-3} - 390u^{-4} + 360u^{-5} - 120u^{-6} du =$

$-(\ln(u) + \dfrac{31}{u} - \dfrac{90}{u^2} + \dfrac{130}{u^3} - \dfrac{90}{u^4} + \dfrac{24}{u^5})|_{1}^{a} = \dfrac{5a^5 - 31a^4 + 90a^3 - 130a^2 + 90a - 24}{a^5} - \ln(a)$

and,

$\int_{a - 1}^{0} f(x) dx = \int_{a - 1}^{0} -\dfrac{-x^6 + 57x^5 - 302x^4 + 302x^3 - 57x^2 + x}{(1 + x)^7} dx =$ $\int_{1}^{a} \dfrac{u^6 - 63u^5 + 602u^4 - 2100u^3 + 3360u^2 - 2520u + 720}{u^7}du =$ $\int_{1}^{a} \dfrac{1}{u} - 63u^{-2} + 602u^{-3} -2100u^{-4} + 3360u^{-5} - 2520u^{-6} + 720u^{-7} du =$ $(\ln(u) + \dfrac{63}{u} - \dfrac{301}{u^2} + \dfrac{700}{u^3} - \dfrac{840}{u^4} + \dfrac{504}{u^5} - \dfrac{120}{u^6})|_{1}^{a} =$ $\ln(a) + \dfrac{63}{a} - \dfrac{301}{a^2} + \dfrac{700}{a^3} - \dfrac{840}{a^4} + \dfrac{504}{a^5} - \dfrac{120}{a^6} - 6 =$ $-(\dfrac{6a^6 - 63a^5 + 301a^4 - 700a^3 + 840a^2 - 504a + 120}{a^6} - \ln(a))$

$\implies$

$\int_{a - 1}^{0} g(x) - f(x) dx = \dfrac{11a^6 - 94a^5 + 391a^4 - 830a^3 + 930a^2 - 528a + 120}{a^6} - 2\ln(a) =$ $\dfrac{11a^6 - 94a^5 - 9609a^4 + 3170a^3 + 330a^2 - 488a + 119}{a^6} - 2\ln(a)$

$\implies \dfrac{10000a^4 - 4000a^3 + 600a^2 - 40a + 1}{a^6} = 0 \implies$ $(10a - 1)^4 = 0 \implies a = \boxed{\dfrac{1}{10}} \implies \int_{\frac{-9}{10}}^{0} g(x) - f(x) dx = \boxed{75708175.605170}$ to six decimal places.

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The area could have been obtained using $f^{*}(x) = \sum_{n = 1}^{\infty} n^6 x^n$ and $g^{*}(x) = \sum_{n = 1}^{\infty} n^5 x^n$ on $[0,1 - a]$ . Note the symmetry about the origin above.