Alternating Sums

Let $x > 1$ .

To Show: If $S_{0}(x) = \sum_{n = 1}^{\infty} \dfrac{1}{x^{n - 1}}$ and for $(1 \leq l \leq k)$ $S_{l}(x) = \sum_{n = 1}^{\infty} \dfrac{n^l}{x^n},$ then

$S_{k + 1}(x) = \sum_{n = 1}^{\infty} \dfrac{n^{k + 1}}{x^n} =$ $(\sum_{l = 0}^{k - 1} \left(\begin{array}{cc} k \\ l \\ \ \end{array}\right) S_{l}(x) S_{k - l}(x)) + \dfrac{1}{x} S_{k}(x) S_{0}(x).$

Let $x > 1$ .

$S_{k + 1} = \sum_{n = 1}^{\infty} \dfrac{n^{k + 1}}{x^n} = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} \dfrac{n^k}{x^n} = \sum_{j = 1}^{\infty} \sum_{n = 0}^{\infty} \dfrac{(j + n)^k}{x^{j + n}} =$

$\sum_{j = 1}^{\infty} (\dfrac{j^k}{x^{j}} \sum_{n = 1}^{\infty} \dfrac{1}{x^{n - 1}} + \left(\begin{array}{cc} k \\ 1 \\ \ \end{array}\right) \dfrac{j^{k - 1}}{x^j} \sum_{n = 1}^{\infty} \dfrac{n}{x^n} + \left(\begin{array}{cc} k \\ 2 \\ \ \end{array}\right) \dfrac{j^{k - 2}}{x^j} \sum_{n = 1}^{\infty} \dfrac{n^2}{x^n} + ... +$ $\left(\begin{array}{cc} k \\ l \\ \ \end{array}\right) \dfrac{j^{k - l}}{x^j} \sum_{n = 1}^{\infty} \dfrac{n^l}{x^n} + ... + \dfrac{1}{x^j} \sum_{n = 1}^{\infty} \dfrac{n^k}{x^n} )$ .

Letting $S_{0}(x) = \sum_{n = 1}^{\infty} \dfrac{1}{x^{n - 1}}$ and for $(1 \leq l \leq k)$ $S_{l}(x) = \sum_{n = 1}^{\infty} \dfrac{n^l}{x^n} \implies$

$S_{k + 1} = S_{0}( x) S_{k}(x) + \left(\begin{array}{cc} k \\ 1 \\ \ \end{array}\right) S_{1}(x) S_{k - 1}(x) + \left(\begin{array}{cc} k \\ 2 \\ \ \end{array}\right) S_{2}(x) S_{k - 2}(x) + ... + \left(\begin{array}{cc} k \\ l \\ \ \end{array}\right) S_{l}(x) S_{k - l}(x) = ... +$

$\dfrac{1}{x} S_{k}(x) S_{0}(x) = \sum_{n = 1}^{\infty} \dfrac{n^{k + 1}}{x^n} =$ $(\sum_{l = 0}^{k - 1} \left(\begin{array}{cc} k \\ l \\ \ \end{array}\right) S_{l}(x) S_{k - l}(x)) + \dfrac{1}{x} S_{k}(x) S_{0}(x)$ .

From the summation it should be clear that $S_{1}(x) = \sum_{n = 1}^{\infty} \dfrac{n}{x^{n - 1}} = \dfrac{1}{x} (S_{0}(x))^2$ .

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Using the above:

Let $x > 1$ .

Clearly the geometric series $S_{0} = \sum_{n = 1}^{\infty} \dfrac{1}{x^{n - 1}}= \dfrac{x}{x - 1}$

For $k = 0: S_{1}(x) = \sum_{n = 1}^{\infty} \dfrac{n}{x^{n - 1}} = \dfrac{1}{x} (S_{0}(x))^2 = \dfrac{1}{x}(\dfrac{x}{x - 1})^2 = \dfrac{x}{(x - 1)^2}$ .

For $k = 1: S_{2}(x) = \sum_{n = 1}^{\infty} \dfrac{n^2}{x^{n}} = S_{0}(x) S_{1}(x) +\dfrac{1}{x} S_{1}(x) S_{0}(x) = \dfrac{x^2 + x}{(x - 1)^3}$

For $k = 2: S_{3}(x) = \sum_{n = 1}^{\infty} \dfrac{n^3}{x^{n}} = S_{0}(x) S_{2}(x) + 2 (S_{1}(x))^2 + \dfrac{1}{x} S_{1}(x) S_{0}(x) = \dfrac{x^3 + 4x^2 + x}{(x - 1)^4}$

For $k = 3: S_{4}(x) = \sum_{n = 1}^{\infty} \dfrac{n^4}{x^{n}} = S_{0}(x) S_{3}(x) + 3 S_{1}(x) S_{2}(x) + 3 S_{2}(x) S_{1}(x) + \dfrac{1}{x} S_{3}(x) S_{0}(x) = \dfrac{x^4 + 11x^3 + 11x^2 + x}{(x - 1)^5}$

$\implies S^*_{4}(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} \dfrac{n^4}{x^n} = -1 * \sum_{n = 1}^{\infty} \dfrac{n^4}{(-x)^n} = -(\dfrac{x^4 - 11x^3 + 11x^2 - x}{-(x + 1)^5}) = \dfrac{x^4 - 11x^3 + 11x^2 - x}{(x + 1)^5}$

Let $u = x + 1 \implies du = dx \implies \int_{2}^{3} S^*_{4}(x) dx = \int_{3}^{4} \dfrac{u^4 - 15u^3 + 50u^2 - 60u + 24}{u^5} du =$ $\int_{3}^{4} \dfrac{1}{u} - 15u^{-2} + 50u^{-3} - 60u^{-4} + 24u^{-5} du = (\ln(u) + \dfrac{15}{u} - \dfrac{25}{u^2} + \dfrac{20}{u^3} - \dfrac{6}{u^4})|_{3}^{4} =$ $\ln(\dfrac{4}{3}) + \dfrac{15}{4} - \dfrac{15}{3} + \dfrac{25}{9} - \dfrac{25}{16} +\dfrac{20}{64} - \dfrac{20}{27} + \dfrac{6}{81} - \dfrac{6}{256} = \ln(\dfrac{4}{3}) - \dfrac{8550}{20736} =$ $\ln(\dfrac{4}{3}) - \dfrac{2 * 3^2 * 5^2 * 19}{2^8 * 3^4} = \ln(\dfrac{2^2}{3}) - \dfrac{5^2 * 19}{2^7 * 3^2} = \ln(\dfrac{c^2}{d}) - \dfrac{a^2 * b}{c^7 * d^2} \implies a + b + c + d = \boxed{29}$

It is a little easier this way, perhaps. If we define $S_k(x) \; = \; \sum_{n=1}^\infty \frac{(-1)^{n+1} n^k}{x^n} \hspace{2cm} x > 1$ (note the slightly different definition for $S_0(x)$ ) then $xS_k'(x) \; = \; -S_{k+1}(x)$ and so, if we define $I_k \; = \; \int_2^3 S_k(x)\,dx$ then integration by parts gives $I_k - I_{k+1} \; = \; \Big[x S_k(x)\Big]_2^3 \; = \; 3S_k(3) - 2S_k(2)$ and hence $I_0 - I_4 \; = \; 3F(3) - 2F(2)$ where $F(x) \; = \; S_0(x) + S_1(x) + S_2(x) + S_3(x)$ Summing the GP that is $S_0(x)$ , and differentiating a few times, we obtain $\begin{aligned} S_0(x) & = \; \frac{1}{x+1} \\ S_1(x) & = \; \frac{x}{(x+1)^2} \\ S_2(x) & = \; \frac{x(x-1)}{(x+1)^3} \\ S_3(x) & = \; \frac{x(x^2 - 4x + 1)}{(x+1)^4} \\ F(x) & = \; \frac{(4x+1)(x^2+1)}{(x+1)^4} \end{aligned}$ for $x > 1$ , and hence $I_4 \; = \; I_0 - 3F(3) + 2F(2) \; = \; \ln\tfrac43 - \tfrac{475}{1152} \; = \; \ln\big(\tfrac{2^2}{3}\big) - \tfrac{5^2 \times 19}{2^7 \times 3^2}$ making the answer $5 + 19 + 2 + 3 = \boxed{29}$ .