Is this a Taylor series? $\ln(x) +1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}$ Can you provide a proof of this identity using only calculus?
$$\ln x + 1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln x)^n}{x}$$
By the way, here is how I arrived at it:
There is string of length $x$ units. Select a point on the string uniformly at random and cut the string at that point. Repeat the process with the string on the left side of the cut until the string you have is shorter than $1$ unit. The problem is to figure out the expected number of cuts. 
Here is how I did it: 
Let $E(x)$ denote the expected number of cuts to be made on a string of length $x$. If $x<1$, clearly, $E(x)=0$. If $x>1$, we have:
\begin{align} E(x) &= 1 + \int_0^x E(u) \cdot \frac{du}{x} \\
                   &= 1 + \frac 1x \int_1^x E(u) \ du
\end{align}  
Multiplying by $x$ and differentiating (applying the Fundamental Theorem of Calculus),
\begin{align} xE'(x) +  E(x) &= 1 + E(x) \\
   \Rightarrow E(x) &= \ln x + C \end{align}
Clearly for $x=1, \ E(x)=1$ thus $E(x) = \ln x + 1$
However, we can also calculate $E(x)$ in a different way:
Let $P(n,\ x)$ denote the probability of exactly $n$ cuts being made on a string of length $x$.
If $x<1$, $P(n,\ x) = 0$; if $n=1$ and $x>1$ $P(n,\ x)=\frac 1x$; if $n>1$ and $x>1$:
\begin{align} P(n,\ x) &= \int_0^x P(n-1,\ u) \cdot \frac{du}{x} \\
                       &= \frac 1x \int_1^x P(n-1,\ u) \ du \end{align}   
I calculated that $P(1,\ x) = \frac 1x,\ P(2,\ x) = \frac{\ln x}{x},\ P(3,\ x) = \frac{(\ln x)^2}{2x},\ P(4,\ x) = \frac{(\ln x)^3}{6x}$
This led me to hypothesize that $P(n,\ x)=\frac{(\ln x)^{n-1}}{x(n-1)!}$, which can be proven by induction:
\begin{align} P(n,\ x) &= \frac 1x \int_1^x P(n-1,\ u) \ du \\
                       &= \frac 1x \int_1^x \frac{(\ln u)^{n-2}}{u(n-2)!} \ du \\                     &= \frac 1{x(n-2)!} \int_1^x (\ln u)^{n-2}\ d(\ln u) \\
                       &= \frac 1{x(n-2)!} \left[\frac {(\ln u)^{n-1}}{n-1}  \right]_1^x \\         &= \frac{(\ln x)^{n-1}}{x(n-1)!} \end{align}
But then $E(x)$ can be written as follows:
\begin{align} E(x) &= \sum_{n=1}^{\infty} n \cdot P(n,\ x)  \\ 
                   &= \sum_{n=1}^{\infty} n \cdot \frac{(\ln x)^{n-1}}{x(n-1)!} \\                 &=  \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x} 
\\ \\
\therefore \ \ln x + 1 &=  \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x} \ \blacksquare \end{align}
So can you prove this result by finding an appropriate Taylor series (this would be especially appreciated) or if not using Taylor series then using just the methods of calculus?

For, $x>0$, $\text{ln}x =z \in (-\infty, \infty)$. 
So, we can write the r.h.s as ${e^{-z}}{\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}$. 
Now, ${\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}$
$=\frac{d (\sum_{n=0}^{\infty} \frac{z^{n+1}}{n!})}{dz}$
$=\frac{d {e^{z}}z}{dz}= e^{z}+ze^{z}$
So, ${e^{-z}}{\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}={e^{-z}}(e^z+ze^{z})=z+1=\text{ln}x+1$. 
