Let $x\in(0,1)$. My goal is to compute this well known almost-geometric series. $$\sum_{k\geq0}(k+1)x^{k}$$
I've computed it using two similar methods, leading to different results. That pineapple brain of mine can't find the error.
method 1 :
\begin{align} \sum_{k\geq0}(k+1)x^{k}=& \sum_{k\geq-1}(k+1)x^{k} && \text{since }k=-1\Rightarrow (-1+1)x^{-1}=0 \\ =&\sum_{\tilde{k}\geq0}\tilde{k}\cdot x^{\tilde{k}-1} && \text{letting } \tilde{k}-1=k\\ =&\sum_{\tilde{k}\geq0}\frac{d}{dx}x^{\tilde{k}}\\ =&\frac{d}{dx}\sum_{\tilde{k}\geq0}x^{\tilde{k}} \end{align}
which is the derivative of an easily-computable geometric series.
method 2 :
\begin{align} \sum_{k\geq0}(k+1)x^{k}=& 1 + \sum_{k\geq1}(k+1)x^{k} && \text{since }k=0\Rightarrow (0+1)x^0=1 \\ =& 1 + \frac{d}{dx}\sum_{k\geq1}x^{k+1}\\ =& 1 + \frac{d}{dx}\sum_{k\geq1}x^{k+1} \\ =& 1 + \frac{d}{dx}\sum_{\tilde{k}\geq0}x^{\tilde{k}} && \text{letting } \tilde{k}-1=k\\ \end{align}
And a $1$ has been created.
Any help would be greatly appreciated.