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

Let $u = x - 1 \implies dx = du \implies \int_{m + 1}^{m + 2} S_{4}(x) dx = \int_{m}^{m + 1} \dfrac{u^4 + 15u^3 + 50u^2 + 60u + 24}{u^5} du =$ $\int_{m}^{m + 1} \dfrac{1}{u} + 15u^{-2} + 50u^{-3} + 60u^{-4} + 24u^{-5} du = (\ln(u) - \dfrac{15}{u} - \dfrac{25}{u^2} - \dfrac{20}{u^3} - \dfrac{6}{u^4})|_{m}^{m + 1} =$

$\ln(\dfrac{m + 1}{m}) + \dfrac{15m^6 + 95m^5 + 230m^4 + 259m^3 + 141m^2 + 44m + 6}{m^4 (m + 1)^4} =$ $\dfrac{m^8 + 4 m^7 + 21 m^6 + 99 m^5 + 231 m^4 + 259 m^3 + 141 m^2 + 44m + 6}{m^4(m + 1)^4} =$ $\dfrac{m^8 + 4m^7 + 6m^6 + 4m^5 + m^4 + 15m^6 + 95m^5 + 230m^4 + 259m^3 + 141m^2 + 44m + 6}{m^4 (m + 1)^4} =$ $1 + \dfrac{15m^6 + 95m^5 + 230m^4 + 259m^3 + 141m^2 + 44m + 6}{m^4 (m + 1)^4} \implies$ $\ln(\dfrac{m + 1}{m}) = 1 \implies \dfrac{m + 1}{m} = e \implies m = \dfrac{1}{e - 1} = \boxed{0.5819767}$ .

Using a method different from that suggested. Let $\displaystyle S_0(x) = \sum_{\color{#3D99F6} n=0}^\infty \frac 1{x^n}$ and $\displaystyle S_k(x) = \sum_{\color{#3D99F6} n=0}^\infty \frac {n^k}{x^n} = \sum_{\color{#D61F06}n=1}^\infty \frac {n^k}{x^n}$ for $k \ge 1$ for $x >1$ . Then we have:

$\begin{aligned} S_k(x) & = \sum_{\color{#3D99F6} n=0}^\infty \frac {n^k}{x^n} = \sum_{\color{#D61F06}n=1}^\infty \frac {n^k}{x^n} = \sum_{\color{#3D99F6} n=0}^\infty \frac {(n+1)^k}{x^{n+1}} = \frac 1x \sum_{n=0}^\infty \frac {\sum_{j=0}^k \binom kj n^j}{x^n} = \frac 1x \sum_{j=0}^k \binom kj \sum_{n=0}^\infty \frac {n^j}{x^n} = \frac 1x \sum_{j=0}^k \binom kj S_j(x) \\ \implies x S_k(x) & = \sum_{j=0}^k \binom kj S_j(x) = S_k(x) +\sum_{j=0}^{k-1} \binom kj S_j(x) \\ \implies S_k(x) & = \frac 1{x-1} \sum_{j=0}^{k-1} \binom kj S_j(x) \end{aligned}$

Now, we have:

$\begin{aligned} S_0 & = \sum_{n=0}^\infty \frac 1{x^n} = \frac x{x-1} \\ S_1 & = \sum_{n=0}^\infty \frac n{x^n} = \frac 1{x-1} \times \frac x{x-1} = \frac x{(x-1)^2} \\ S_2 & = \sum_{n=0}^\infty \frac {n^2}{x^n} = \frac 1{x-1} \left(\frac x{x-1} +\frac {2x}{(x-1)^2}\right) = \frac {x^2+x}{(x-1)^3} \\ S_3 & = \sum_{n=0}^\infty \frac {n^3}{x^n} = \frac 1{x-1} \left(\frac x{x-1} +\frac {2x}{(x-1)^2} + \frac {3(x^2+x)}{(x-1)^3} \right) = \frac {x^3+4x^2+x}{(x-1)^4} \\ S_4 & = \sum_{n=0}^\infty \frac {n^4}{x^n} = \frac 1{x-1} \left(\frac x{x-1} +\frac {4x}{(x-1)^2} + \frac {6(x^2+x)}{(x-1)^3} + \frac {4(x^3+4x^2+x)}{(x-1)^4} \right) = \frac {x^4 + 11x^3+11x^2+x}{(x-1)^5} \end{aligned}$

Therefore, we have:

$\begin{aligned} I & = \int_{m+1}^{m+2} \sum_{n=0}^\infty \frac {n^4}{x^n} dx \\ & = \int_{m+1}^{m+2} \frac {x^4 + 11x^3+11x^2+x}{(x-1)^5} dx \quad \quad \quad \quad \small \color{#3D99F6} \text{Let }u = x - 1 \implies x = u +1, \ du = dx \\ & = \int_m^{m+1} \frac {u^4+15u^3+50u^2+60u}{u^5} du \\ & = \int_m^{m+1} \left(\frac 1u+\frac {15}{u^2}+\frac {50}{u^3}+\frac {60}{u^4} + \frac {24}{u^5} \right) du \\ & = \ln u - \frac {15}u- \frac {25}{u^2}- \frac {20}{u^3} - \frac 6{u^4}\ \bigg|_m^{m+1} \\ & = \ln \left(\frac {m+1}m\right) - \frac {15(m+1)^3+25(m+1)^2+20(m+1)+6}{(m+1)^4} + \frac {15m^3+25m^2+20m+6}{m^4} \\ & = \ln \left(\frac {m+1}m\right) + \frac {15m^6+95m^5+230m^4+259m^3+141x^2+44m+6}{m^4(m+1)^4} \\ & = \ln \left(\frac {m+1}m\right) + \frac {{\color{#3D99F6}x^8+4m^7+21m^6+99m^5+231m^4}+259m^3+141x^2+44m+6}{m^4(m+1)^4} - \frac {\color{#D61F06}x^8+4m^7+6m^6+4m^5+1m^4}{m^4(m+1)^4} \\ & = {\color{#D61F06}\ln \left(\frac {m+1}m\right)} + \frac {x^8+4m^7+21m^6+99m^5+231m^4+259m^3+141x^2+44m+6}{m^4(m+1)^4} \color{#D61F06}-1 \end{aligned}$

This implies that:

$\begin{aligned} \ln \left(\frac {m+1}m\right) & = 1 \\ \frac {m+1}m & = e \\ \implies m & = \frac 1{e-1} \approx \boxed{0.5819767}\end{aligned}$