Shifting index while calculating expected value 
Let be $X$ a random variable, $X:\Omega \to \mathbb{N}\cup \{0\}$. Let's assume the expected value $\mathbb{E}(X)=\sum\limits_{k=0}^{\infty}kP(X= k)$ does exist.

Is it legit to replace $k$ by $(k+1)$:
$\mathbb{E}(X)=\sum\limits_{k=0}^{\infty}kP(X= k)=\sum\limits_{k=0}^{\infty}(k+1)P(X= k+1)$ ?
If you consider a finite sum then you have to reduce the last index by $1$, e.g. $\sum\limits_{k=0}^{5}kP(X= k)=\sum\limits_{k=0}^{4}(k+1)P(X= k+1)$.
But does it change anything if you have an infinite sum/series?
 A: It is legitimate. $ \sum\limits_{k=0}^{\infty} ka_k= \sum\limits_{k=0}^{\infty} (k+1)a_{k+1}$ whenever the series converges. To prove this write down the partial sums ans observe that $na_n \to 0$ as $ n \to \infty$ so the difference between the partial sums tends to $0$.
A: 
If you consider a finite sum then you have to reduce the last index by $1$

Technically, you must reduce both indices, however since the first term vanishes when $k=0$, then this happens:
$$\begin{align}\sum_{k=0}^n k\,\mathsf P(X\,{=}\,k)~&=~0+\sum_{k=1}^n k\,\mathsf P(X\,{=}\,k)\\[1ex]&=~\sum_{k=0}^{n-1}(k+1)\,\mathsf P(X\,{=}\,k\,{+}\,1)\end{align}$$
In the infinite case, as long as the series is convergent then we don't have to worry about doing anything to the infinite upper limit:$$\begin{align}\sum_{k=0}^\infty k\,\mathsf P(X\,{=}\,k)~&=~\lim_{n\to\infty}\sum_{k=1}^n k\,\mathsf P(X\,{=}\,k)\\[1ex]&=~\lim_{n\to\infty}\sum_{k=0}^{n-1}(k+1)\,\mathsf P(X\,{=}\,k\,{+}\,1)\\[1ex]&=~\lim_{n-1\to\infty}\sum_{k=0}^{n-1}(k+1)\,\mathsf P(X\,{=}\,k\,{+}\,1)\\[1ex]&=~\sum_{k=0}^{\infty}(k+1)\,\mathsf P(X\,{=}\,k\,{+}\,1)\end{align}$$
