Where does it come $\left[\sum_{n=0}^{\infty}x^n\right]' = \sum_{n=0}^{\infty}(n+1)x^n$

Question

Where does it come $$\left[\sum_{n=0}^{\infty}x^n\right]' = \sum_{n=0}^{\infty}(n+1)x^n$$?

Answer 1

Differentiate the given sum termwise and you will get

$$\left[\sum_{n=0}^{\infty}x^n\right]' = \sum_{n=0}^{\infty}\left[x^n\right]'=\sum_{n=0}^{\infty}nx^{n-1}$$

Hence the first term of this sum equals zero we can let start the series at $n=1$ as well. Plugging in this change and applying an index shift afterwards leads to

$$\sum_{n=0}^{\infty}nx^{n-1}=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^{n}$$

Answer 2

$(x^n)' = n x^{n-1}$, with care given to interchange of limits.

Answer 3

Disclaimer

I will try to give a clearer answer, but this is essentially the same as the one provided by mrtaurho.

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Look at this:

$$\left[\sum_{n=0}^{\infty}x^n\right]' = \left[x^0 + \sum_{n=1}^{\infty}x^n\right]' = \left[1 + \sum_{n=1}^{\infty}x^n\right]' = 0 + \sum_{n=1}^{\infty}nx^{n-1} = \sum_{n=1}^{\infty}nx^{n-1}.$$

Now, let's pose $m = n-1$. It is clear that:

1. $m=0$ when $n = 1$;
2. $m=\infty$ when $n = \infty$;
3. $n = m + 1$.

Therefore:

$$\left[\sum_{n=0}^{\infty}x^n\right]' = \sum_{m=0}^{\infty}(m+1)x^{m}.$$

At this point, you are free to rename $n = m$ (but this "$n$" is different from the "$n$" used at the very beginning!!! pay attention on this point!!!), thus obtaining the following:

$$\left[\sum_{n=0}^{\infty}x^n\right]' = \sum_{n=0}^{\infty}(n+1)x^{n}.$$