Laurent series of $\left(1+\frac{1}{x}\right)^x$ about $x=\infty$/MacLaurin series of $\left(1+x\right)^\frac{1}{x}$

Question

For those who do not know, a Laurent series is like a Taylor series, but with terms that can have exponents of negative degree. I was wondering how I can find the Laurent series of $$f(x)=\left(1+\frac{1}{x}\right)^x$$ about $x=\infty$. Entering into WolframAlpha, we get that $$\left(1+\frac{1}{x}\right)^x = e-\frac{e}{2x}+\frac{11e}{24x^2}+... = e\sum_{n=0}^\infty a_nx^{-n}$$ where the $a_n$ are very likely all rational numbers. I want to find the values of $a_n$.

To hopefully make this simpler, this is equivalent to finding the values of $a_n$ in the MacLaurin series of $$(1+x)^\frac{1}{x} = e-\frac{e}{2}x+\frac{11e}{24}x^2+... = e\sum_{n=0}^\infty a_nx^n$$

I honestly do not know what to do from here, as taking the $n$th derivative just makes it more complicated.

Answer 1

Let $x=-\frac 1t$ and consider the series expansion of $$\frac 1 e (1-t)^{-1/t}$$ built at $t=0$. It will write $$\frac 1 e (1-t)^{-1/t}=\sum_{n=0}^\infty \frac {a_n}{b_n} t^n$$ for which the first terms are

• for $a_n$

$$\{1,1,11,7,2447,959,238043,67223,559440199,123377159,29128857391\}$$

• for $b_n$

$$\{1,2,24,16,5760,2304,580608,165888,1393459200,309657600,73574645760\}$$

These correspond respectively to sequences $A055505$ and $A055535$ on $OEIS$.

Looking at the comments in the linked $OEIS$ pages, for the series expansion of $$ (1+t)^{1/t}=\sum_{n=0}^\infty {c_n} t^n$$ we have $$\color{blue}{c_n=\sum _{k=0}^n \frac{ \Gamma (n+1-k,-1)}{ (n+k)! \,(n-k)!}\,S_{k+n}^{(k)}}$$ where appear Stirling numbers of the first kind.

These are your coefficients.

Answer 2

All of this is purely formal, be my guest and make it rigorous.

My suggestion would be the following brute force approach. However, I don't know whether this is useful for you. We can write $$ (1+x)^{1/x} = e^{\frac{1}{x} \ln(1+x)} $$ Furthermore, we have $$ \frac{1}{x} \ln(1+x) = \sum_{n\geq 0} \frac{(-1)^n}{n+1} x^n.$$ When we plug this into the expansion of the exponential we get using the idea of Monadologie $$ (1+x)^{1/x} = \sum_{k\geq 0} \frac{1}{k!} \left( \sum_{n\geq 0} \frac{(-1)^n}{n+1} x^n \right)^k =e \sum_{k\geq 0} \frac{1}{k!} \left( \sum_{n\geq 1} \frac{(-1)^n}{n+1} x^n \right)^k = \sum_{k\geq 0} e \frac{1}{k!} \sum_{N\geq 0} \left( \sum_{1\leq \lambda_1, \dots, \lambda_k \ : \ \lambda_1 + \dots+ \lambda_k =N} \prod_{j=1}^k \frac{(-1)^{\lambda_j}}{\lambda_j+1} \right) x^N = \sum_{N\geq 0} (-1)^N e \left( \sum_{k\geq 0} \frac{1}{k!} \sum_{1\leq \lambda_1, \dots, \lambda_k \ : \ \lambda_1 + \dots+ \lambda_k =N} \prod_{j=1}^k \frac{1}{\lambda_j+1} \right) x^N $$

Answer 3

Note: Nothing original here.

$\begin{array}\\ f(x) &=\left(1+\frac{1}{x}\right)^x\\ &=\exp(x\ln(1+\frac1{x}))\\ &=\exp(x\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{nx^n})\\ &=\exp(\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{nx^{n-1}})\\ &=\exp(\sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{(n+1)x^{n}})\\ &=\exp(1+\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{(n+1)x^{n}})\\ &=e\exp(\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{(n+1)x^{n}})\\ \end{array} $

At this point, you want to get $\exp(g(x))$ for a power series $g(x) =\sum_{n=1}^{\infty} g_nx^n $.

Here is a standard way to do this:

Let $h(x) = \exp(g(x)) $ where $g(0) = 0$.

We will get a recurrence for the coefficients of $h(x)$.

Then $h'(x) =(\exp(g(x)))' =g'(x)\exp(g(x)) =g'(x) h(x) $.

Let $h(x) =\sum_{n=0}^{\infty} h_n x^n $ and $g(x) =\sum_{n=0}^{\infty} g_n x^n $ with $g_0=0$.

Then $h'(x) =\sum_{n=0}^{\infty} nh_n x^{n-1} =\sum_{n=0}^{\infty} (n+1)h_{n+1} x^{n} $ and $g'(x) =\sum_{n=0}^{\infty} ng_n x^{n-1} =\sum_{n=0}^{\infty} (n+1)g_{n+1} x^{n} $.

Therefore, from $h'(x) =g'(x) h(x) $ we have

$\begin{array}\\ \sum_{n=0}^{\infty} (n+1)h_{n+1} x^{n} &=\sum_{n=0}^{\infty} (n+1)g_{n+1} x^{n} \sum_{m=0}^{\infty} h_m x^m\\ &=\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} (n+1)g_{n+1}h_m x^{n+m}\\ &=\sum_{n=0}^{\infty} x^n \sum_{k=0}^{n} (k+1)g_{k+1}h_{n-k}\\ \text{so}\\ (n+1)h_{n+1} &=\sum_{k=0}^{n} (k+1)g_{k+1}h_{n-k}\\ \text{or}\\ h_{n+1} &=\dfrac1{n+1}\sum_{k=0}^{n} (k+1)g_{k+1}h_{n-k}\\ &=\dfrac1{n+1}\sum_{k=0}^{n} (n-k+1)g_{n-k+1}h_{k}\\ \end{array} $

To initialize, $h_0 =h(0) =\exp(g(0)) =\exp(g(0)) =1 $.

We have $g_n =(0, -\frac12, \frac13, -\frac14, ... )$, so

$n=0: h_1 =g_1h_0 =-\dfrac12 $.

$n=1: h_2 =\dfrac1{2}(2g_2h_0+g_1h_1) =\dfrac12(2(\frac13)-\frac12(-\frac12)) =\dfrac12(\frac23+\frac14) =\dfrac{11}{24} $.

$n=2: h_3 =\dfrac1{3}(3g_3h_0+2g_2h_1+g_1h_2) =\dfrac1{3}(3(-\frac14)+2(\frac13)(-\frac12)+(-\frac12)\frac{11}{24})\\ =\dfrac1{3}(-\frac34-\frac13-\frac{11}{48}) =\dfrac1{3}\dfrac{-3\cdot 12-16-11}{48}) =\dfrac{-63}{3\cdot 48} =\dfrac{-7}{16} $.

And so on.

I don't know of a general expression for the $h_n$ that doesn't involve nested summations.