how to continue?

[lost whale]

l'agit

[lost whale]

l'agit

[turbid sigil]

Do you agree that
$$\mathbb{E}[X] = \sum_{n=1}^{\infty} np(1-p)^{n-1}$$

[lost whale]

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[turbid sigil]

which is just
$$\mathbb{E}[X] = p\sum_{n=1}^{\infty}n(1-p)^{n-1}$$

[lost whale]

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[turbid sigil]

and I will give you a hint that $\dfrac{d}{dp} (1-p)^{n} = n(1-p)^{n-1}$ and $\dfrac{d}{dp} \sum_{i=1}^\infty f(p) = \sum_{i=1}^\infty \dfrac{d}{dp} f(p)$

[lost whale]

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