It is desirable to be able to differentiate and multiply power series. The following theorem says you can differentiate power series in the most natural way on the interval of convergence, just as you would differentiate a polynomial. This theorem may seem obvious, but it is a serious mistake to think this. You usually cannot differentiate an infinite series whose terms are functions even if the functions are themselves polynomials. The following is special and pertains to power series. It is another example of the interchange of two limits, in this case, the limit involved in taking the derivative and the limit of the sequence of finite sums.
When you formally differentiate a series term by term, the result is called the derived series.
Theorem 8.2.1 Let ∑ _{n=0}^{∞}a_{n}
 (8.2) 
for
 (8.3) 
and this new differentiated power series, the derived series, has radius of convergence equal to R.
Proof: Let
Then for all n large enough,
Let δ be small enough that if

By the mean value theorem, there exists θ_{nh} ∈

By the mean value theorem again, there exists α_{nh} ∈

The second series is the derived series. Consider the first.

and so the series converges. Hence, letting h → 0 yields the desired result that

As an immediate corollary, it is possible to characterize the coefficients of a Taylor series.
Corollary 8.2.2 Let ∑ _{n=0}^{∞}a_{n}
 (8.4) 
Then
 (8.5) 
Proof: From 8.4, f

Now let x = a and obtain that f^{′}

let x = a in this equation and obtain a_{2} = f^{′′}
This also shows the coefficients of a Taylor series are unique. That is, if

for all x in some open set containing a, then a_{k} = b_{k} for all k.
Example 8.2.3 Find the sum ∑ _{k=1}^{∞}k2^{−k}.
It may not be obvious what this sum equals but with the above theorem it is easy to find. From the formula for the sum of a geometric series,

whenever

and so if you multiply both sides by 2^{−1},

The above theorem shows that a power series is infinitely differentiable. Does it go the other way? That is, if the function has infinitely many continuous derivatives, is it correctly represented as a power series? The answer is no. See Problem 6 on Page 482 for an example. In fact, this is an important example and distinction. The modern theory of partial differential equations is built on just such functions which have many derivatives but no power series.