Confusion about proof involving sums and binomial coefficients

Question

In the below proof, how is the second line derived from the first?

The first term in the second line has n replaced by (n+1), I cant see why, specially, because this change is not made in the second term.

I understand the rest of the proof but not this part.

What is the manipulation that is being performed?

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Answer 1

It's a reindexing of the sum. The general form is $$ \sum_{n=1}^\infty a_n x^{n-1} = \sum_{n=0}^\infty a_{n+1} x^{n} \tag{1} $$ by a change of variable "$n\leadsto n+1$" (are you comfortable with this?).

To see how it is applied here:

• First sum: $$ \sum_{n=1}^\infty \underbrace{n\binom{\alpha}{n}}_{a_n} x^{n-1} = \sum_{n=0}^\infty \underbrace{(n+1)\binom{\alpha}{n+1}}_{a_{n+1}} x^{n}\,. $$

However, in the second sum this is not used. Instead, what is used is simply that $$ \sum_{n=1}^\infty n\binom{\alpha}{n} x^{n} = \sum_{n=1}^\infty n\binom{\alpha}{n} x^{n} + \underbrace{0\cdot \binom{\alpha}{0} x^{0}}_{=0} = \sum_{n=0}^\infty n\binom{\alpha}{n}x^{n}\,, $$ i.e., that adding the term "$a_0 x^0$" to the sum does not change the value here, as $a_0=0$.

Answer 2

Suppose we set $n=N+1$. Since $n$ starts at $n=1$, $N$ starts at $N=0$, and we have $$\sum_{n\ge 1}n\binom\alpha nx^{n-1}=\sum_{N\ge 0}(N+1)\binom\alpha {N+1}x^{N}.$$ Then set $N=n$, as they're dummy variables.