Any recommendations on how to solve this using a power series $x^{x + 1} = (x + 1)^x$?

Question

Any recommendations on how to solve this using a power series:

$$ x^{x+1}=(x+1)^x $$

Figured out you could not do it algebraically so decided to think outside the box.

Answer 1

It's actually not that difficult to get the power series. Using the fact that any number $a^x$ is equal to $e^{x\ln a}$ we can express the power series as $\sum\limits_{n=0}^{\infty}\frac{(x\ln a)^n}{n!}$ from the power series for $e^x$. If a is another variable say $x$ then we just replace it with x. So here we have a power series for both the expressions given

Answer 2

Note that

$$x^{x+1}=x^x\cdot x=(x+1)^x$$

Divide both sides by $x^x$ to get

$$x=\left(1+\frac1x\right)^x$$

Now from here I'll tell you I don't believe there is a closed form, but we can do some quick fixed-point iteration:

$$x_{n+1}=\left(1+\frac1{x_n}\right)^{x_n}$$

With $x_0=2.3$, we get

$x_1=2.2940772541106$

$x_2=2.2932879508444$

$x_3=2.2931825401271$

$x_4=2.2931684586440$

$x_5=2.2931665774723$

Which is the first few digits of the solution.