It starts with an innocent power series

In this post I play around with some blogdown features. For that a crazy proof is resurrected. The sum of a funny, innocent, yet infinite power series is derived. It is very likely, that someone else did this before. Anyway, it is sufficient to brag around with some math and to include an R ggplot.

We are talking about this sweetie:


I just had wondered if there was a more handsome form of its sum value. This thing is a little bit different from an ordinary power series. The variable \(x\) normally does not like to live in the exponent of the coefficient, except in this sweet expression..

The equation we will get: \[\sum_{n=0}^{\infty}\frac{n^a}{a^n}=\sum_{k=1}^{a}k!\cdot S_2(a,k)\cdot\frac{a}{(a-1)^{k+1}}\] To get there, a Lemma is required.

Lemma 1: For the \(n\)-times differentiable function \(A(x)\), \(n>0\), and the differential operator \({\mathrm{D}}=\frac{\mathrm{d}}{\mathrm{d}x}\) there is the equation: \[(x{\mathrm{D}})^n A(x) = \sum_{k=1}^n S_2(n,k)\cdot x^k\cdot {\mathrm{D}}^k A(x)\] where \(S_2(n,k)\) is the Stirling Number of the second kind. The Stirling Number of the second kind can be recursively defined as:

\[S_2(n,k) = S_2(n-1,k-1) + k\cdot S_2(n-1,k)\]

while \(S_2(n,0)=0\) for \(n>0\) and \(S_2(n,m)=0\) for \(m>n\).

Proof: We prove the claim by induction on \(n\). For \(n=1\) it is easy to see that

\[x{\mathrm{D}}A(x) = \sum_{k=1}^{1}S_2(1,k)\cdot x^k {\mathrm{D}}^k A(x)\,,\] where \(S_2(1,1)=1\). This starts the induction. The inductive hypothesis for \(n-1\) states:

\[(x{\mathrm{D}})^{n-1} A(x) = \sum_{k=1}^{n-1} S_2(n-1,k)\cdot x^k\cdot {\mathrm{D}}^k A(x)\]

Hence it holds: \[\begin{aligned} (x{\mathrm{D}})^n A(x) &= x{\mathrm{D}}\left( (x{\mathrm{D}})^{n-1} A(x) \right) \\ &= x{\mathrm{D}}\left( \sum_{k=1}^{n-1} S_2(n-1,k)\cdot x^k\cdot {\mathrm{D}}^k A(x) \right)\\ &=\sum_{k=1}^{n-1}k\cdot S_2(n-1,k)x^k{\mathrm{D}}^kA(x) + \sum_{k=1}^{n-1}S(n-1,k)x^{k+1}{\mathrm{D}}^{k+1}A(x)\\ &=\sum_{k=1}^{n-1}k\cdot S_2(n-1,k)x^k{\mathrm{D}}^kA(x) + \sum_{k=2}^{n}S(n-1,k-1)x^{k}{\mathrm{D}}^{k}A(x)\\\end{aligned}\]

The indices can be extended as follows, because that Stirling Numbers are zeros: \[\begin{aligned} (x{\mathrm{D}})^n A(x) &= \sum_{k=1}^{n}k\cdot S_2(n-1,k)x^k{\mathrm{D}}^kA(x) + \sum_{k=1}^{n}S(n-1,k-1)x^{k}{\mathrm{D}}^{k}A(x)\\ &=\sum_{k=1}^{n}\left( k\cdot S_2(n-1,k)+S_2(n-1,k-1) \right) x^k {\mathrm{D}}^k A(x)\\ &=\sum_{k=1}^{n} S_2(n,k) x^k {\mathrm{D}}^k A(x)\end{aligned}\]

Theorem: For a given \(a>1\) the sum of the following power series can be calculated: \[\sum_{n=0}^{\infty}\frac{n^a}{a^n}=\sum_{n=0}^{\infty}n^a\left(\frac{1}{a}\right)^n=\sum_{k=1}^{a}k!\cdot S_2(a,k)\cdot\frac{a}{(a-1)^{k+1}}\]

Proof: As power series are differentiable on the interior of the domain of convergence one can write, if \(|x|<1\): \[\begin{aligned} \sum_{n=0}^{\infty}n^ax^n &= (x{\mathrm{D}})^a \sum_{n=0}^{\infty}x^n\\ &= (x{\mathrm{D}})^a \frac{1}{1-x}\end{aligned}\]

By Lemma [1] the result can be computed as: \[\begin{aligned} (x{\mathrm{D}})^a \frac{1}{1-x} &= \sum_{k=1}^a S_2(a,k)\cdot x^k\cdot {\mathrm{D}}^k \frac{1}{1-x}\\ &=\sum_{k=1}^a S_2(a,k)\cdot k!\cdot \frac{x^k}{(1-x)^{k+1}}\end{aligned}\] With \(x=\frac{1}{a}\) the result can be written as follows: \[\begin{aligned} \sum_{n=0}^{\infty}\frac{n^a}{a^n} &= \sum_{k=1}^a k!\cdot S_2(a,k)\cdot \frac{a^{-k}}{\left(1-\frac{1}{a}\right)^{k+1}}\\ &=\sum_{k=1}^{a}k!\cdot S_2(a,k)\cdot\frac{a}{(a-1)^{k+1}}\end{aligned}\]

Example: For \(a=2\) the sum is \(\sum_{n=0}^{\infty}\frac{n^2}{2^n}=1!\frac{4}{2}+2!\frac{8}{4}=6\). With \(a=3\) you will obtain \(\frac{66}{16}\).

In Figure 1 you can see the behaviour of the sum approximation.

sumit <- function(len,a) {
    tsum <- 0
    for(n in seq(0,len-1)) {
        tsum <- tsum + n**a / a**n
sumdir <- function(a) {
    tsum <- 0
    for(ka in seq(1,a)) {
        tsum <- tsum + factorial(ka) * Stirling2(a,ka,method="direct")*a/(a-1)**(ka+1)
sumit_to_frame <- function(lengths,a) {
    dfp <- data.frame(x=lengths, y=sapply(lengths,sumit,a=a), a=as.integer(a))
    dfp$c <- as.numeric(sumdir(a))
lens <- seq(1,25)
dfp <-
    sumit_to_frame(lens, 2) %>%
    bind_rows( sumit_to_frame(lens, 3) ) %>%
    bind_rows( sumit_to_frame(lens, 4) ) %>%
    bind_rows( sumit_to_frame(lens, 5) ) 
dfp$a <- as.factor(dfp$a)

my_theme <-  theme_bw() + theme(axis.title.x = element_text(size=18),
                                axis.title.y = element_text(size=18),
                                axis.text.x = element_text(size=14),
                                axis.text.y = element_text(size=14),
                                plot.margin = unit(c(8,10,1,1), "pt") # required otherwise labels are clipped in pdf output
my_theme <- my_theme + theme(#legend.title = element_blank(),
                           legend.title = element_text(size=16, face="bold"),
                           legend.text = element_text( size = 16),
                           legend.background = element_rect(colour = 'white', fill = 'white', size = 0., linetype='dashed'),
                           legend.key = element_rect(colour = 'white', fill = 'white', size = 0., linetype='dashed'),
                           legend.key.width = unit(1.1, "cm"),
                           panel.grid.major = element_line(color="#444444", size=0.5, linetype=3),
                           panel.grid.minor = element_line(color="#555555", size=0.5, linetype=3)

gg <- ggplot(dfp, aes(x,colour=a)) + geom_line(aes(y=y),size=1.2) + geom_line(aes(y=c),linetype="dashed",size=1.0) + geom_point(aes(y=y),size=2.5)
gg <- gg + my_theme
gg <- gg + xlab("iterations N")
gg <- gg + ylab(expression(sum(frac(n**a,a**n),n==0,N)))
Sum approximation for different $a$ and iterations $N$.

Figure 1: Sum approximation for different \(a\) and iterations \(N\).