Question : If $V_n=\frac{d^n}{dx^n}(x^n \log x)$, show that $V_n=nV_{n-1}+(n-1)!$
Hence show that $$V_n=n! (\log x + 1 + \frac{1}{2}+\frac{1}{3}+\dot{} \dot{} \dot{}+\frac{1}{n})$$
What I have managed to do so far:
I have found that $V_{n+1}=\frac{n!}{x}$ but I cannot use it further to answer the questions.
PS: Here $V_n$ is $n^{th}$ derivative of $V$
Can someone kindly guide me on how to pursue further in this problem?
Just compute one of the derivatives in the definition of $V_n$ to get the recursion.
$$V_n = \frac{d^n}{dx^n} (x^n \log x) = \frac{d^{n-1}}{dx^{n-1}} (n x^{n-1} \log x + x^{n-1}) = n V_{n-1} + (n-1)!$$
Then, applying this recursion repeatedly yields \begin{align} V_n &= nV_{n-1} + (n-1)!\\ &= n[(n-1) V_{n-2} + (n-2)!] + (n-1)! \\ &= n(n-1) V_{n-2} + n! \left(\frac{1}{n-1} + \frac{1}{n}\right) \\ &\;\vdots\\ &= n! V_0 + n!\left(1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right), \end{align} where $V_0= \log x$.
