I'm trying to derive the time-dependent Schrödinger's equation (TDSE).
Starting with the definition of the Unitary time evolution operator, $$\hat U(t)=e^{-i\hat H t/\hbar}$$ where $\hat H$ is the Hamiltonian operator.
The Taylor series expansion for an exponentiated operator, $\hat X$ is $$e^{\hat X}=\sum_{n=0}^\infty\frac{\hat X^n}{n!}$$
Therefore, $$\hat U(t)=e^{-i\hat H t/\hbar}=\sum_{n=0}^\infty\left(\frac{-i\hat H t}{\hbar}\right)^n\frac{1}{n!}\tag{1}$$ Differentiating $(1)$ with respect to time and multiplying both sides by $i\hbar$, $$\begin{align}i\hbar\frac{d\hat U(t)}{dt}=i\hbar\frac{d \left(e^{-i\hat H t/\hbar}\right)}{dt}&=\sum_{n=0}^\infty i\hbar \left(\frac{-i}{\hbar}\right)^n\frac{n}{n!}t^{n-1}{\hat H}^n\tag{2}\\&=\sum_{n=0}^\infty i\hbar \left(\frac{-i}{\hbar}\right)^n\frac{t^{n-1}{\hat H}^n}{(n-1)!}\tag{3}\end{align}$$
In the last equality I simply used that $\frac{n}{n!}=\frac{1}{(n-1)!}$
As can be seen from the expression in $(2)$, the $n=0$ term is zero, so I need a summation starting from $n=1$, to do this, I let $k=n+1$
I purposely chose this new index so that when $n=0$, $k=1$, thus rewriting the summation in $(3)$ yields, $$i\hbar\frac{d\,\hat U(t)}{dt}=\sum_{k=1}^\infty i\hbar \left(\frac{-i}{\hbar}\right)^{k-1}\frac{t^{k-1-1}{\hat H}^{k-1}}{(k-2)!}=\sum_{k=1}^\infty i\hbar \left(\frac{-i}{\hbar}\right)^{k-1}\frac{t^{k-2}{\hat H}^{k-1}}{(k-2)!}\tag{?}$$
When I tried to shift the summation index in this expression all I was doing was following Jason Rose's video on changing the index of summation which can be found here.
Now here is the problem, when the lecturer shifts the index in equation $(3)$ in lectures and the handwritten notes the lecturer writes $n\to n+1$, $$\fbox{$\begin{align}i\hbar\frac{d\,\hat U(t)}{dt}&=\sum_{n=0}^\infty i\hbar \left(\frac{-i}{\hbar}\right)^{n+1}\frac{t^{n}{\hat H}^{n+1}}{n!}\\&=\sum_{n=0}^\infty\left(\frac{-i}{\hbar}\right)^n\frac{t^{n}{\hat H}^{n+1}}{n!}\\&=\hat H \hat U(t)\end{align}$}$$
In going from the first to the second equality in the box above, $i\hbar\left(\frac{-i}{\hbar}\right)^{n+1}=\left(\frac{-i}{\hbar}\right)^n$, and in going from the second to the third equality, the final expression only differs from equation $(1)$ by a single factor of $\hat H$.
Specializing the equation in the box above to the case of the Schrödinger picture, the states, $\lvert \psi(t)\rangle$ evolve in time, and so
$$i\hbar\frac{d \lvert\psi(t)\rangle}{dt}=\hat H \lvert \psi(t)\rangle\qquad\blacksquare\tag{TDSE}$$
I don't like the notation $n\to n+1$, as I think this leads to unnecessary confusion that can easily be avoided and I prefer to write $k=n+1$.
But with the exception of the letter $n$ in $n=n+1$ being replaced by $k=n+1$, this logic is identical to that given by the lecturer and so should give exactly the same answer.
So why does the summation index relabelling not work in my equation, $(\mathrm{?})$, despite the fact that I am doing the exact same thing as the lecturer (except with $n$ replaced by $k$)?
For $n\ge0$, define$$u_n:=i\hbar \left(\frac{-i}{\hbar}\right)^n\frac{n}{n!}t^{n-1}{\hat H}^n$$so $u_0=0$; for $n\ge1$, define$$v_n:=i\hbar \left(\frac{-i}{\hbar}\right)^n\frac{t^{n-1}{\hat H}^n}{(n-1)!}$$so $u_n=v_n$. Of course, $v_0$ is undefined. The correct manipulation is$$\sum_{n\ge0}u_n=\sum_{n\ge1}u_n=\sum_{n\ge1}v_n=\sum_{n\ge0}v_{n+1},$$or equivalently$$\sum_{k\ge1}u_{k-1}=\sum_{k\ge2}u_{k-1}=\sum_{k\ge2}v_{k-1}=\sum_{k\ge1}v_k.$$What you've attempted is more like $\sum_{n\ge0}u_n=\sum_{n\ge0}v_n$, which doesn't work regardless of your choice of letters.