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$$-\sum_{x=1}^{\infty} x(1-\zeta)^{x-1}=\sum_{x=1}^{\infty} \frac{d}{d\zeta}(1-\zeta)^x\stackrel{?}{=}\frac{d}{d\zeta}\sum_{x=1}^{\infty} (1-\zeta)^x$$

Can anyone explain if switching the derivative and sum is allowed here and if it is give me a rough explanation/intuition as to why it is true?

Thanks.

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  • $\begingroup$ Yes, it is correct. $\endgroup$ Feb 1, 2017 at 17:03
  • $\begingroup$ Could you give me some rough idea (I'm not after pure rigour at the moment) why this is so? $\endgroup$
    – Ryan S
    Feb 1, 2017 at 17:04

2 Answers 2

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The gnomic answer is that the derivative is a linear operator on functions.

But you can really see this from the definition of derivative:

$$(f+g)'(x) = \lim_{h\to 0} \frac{(f+g)(x+h)-(f+g)(x)}h = $$ $$ = \lim_{h\to 0} \frac{(f(x+h)+g(x+h)-f(x)-g(x))(x)}h $$ $$ = \lim_{h\to 0} \left[\frac{f(x+h)-f(x)}h +\frac{g(x+h)-f(x)}h \right]$$ $$ = \lim_{h\to 0} \frac{f(x+h)-f(x)}h +\lim_{h\to 0}\frac{g(x+h)-f(x)}h$$ $$ = f'(x) + g'(x)$$

And if this is true for the sum of two functions, it is true for an arbitrary number of terms by finite induction.

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Your work is correct. The main intuition behind this is that the variable on which the summation is being carried out and the variable with respect to which the sum is being differentiated are different, or in other words, independent.

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