Contradictory results when using formula for infinite geometric progression

Question

Let $$f(x)=\sum_{i=1}^{\infty} (-1)^{i-1}\frac{x^i}{i}$$

Then the simple question is to find out $f'(1)$

I did it in two ways--

Differentiating f(x) we get $$f'(x)=\sum_{i=1}^{\infty} (-1)^{i-1}x^{i-1}$$.Putting $x=1$ we see that all terms cancels out.Hence $f'(1)=0$

Now $f'(x)$ can also be written using the formula for infinite G.P. as

$$f'(x)=\frac{1}{1+x}$$

Putting in $x=1$ we get $f'(x)=1/2$.

How are these two results contradictory.Can someone explain which one is wrong and why?Thanks.

Answer 1

Putting $x=1$ we see that all terms cancels out

Or do they? Have a close look at the sequence under the series for $x=1$:

$$1, -1, 1, -1, 1, -1, \ldots$$

So the sequence of partial sums is

$$1, 0, 1, 0, 1, 0, \ldots$$

This sequence is not convergent, i.e. the series is not convergent.

The series $\sum (-1)^{n}x^n$ is only convergent for $|x|<1$. For other $x$ you cannot conclude anything.

Answer 2

Here the limit function $$f(x)=\log(1+x)$$ on $(-1,1)$ and we can differentiate a series of function $$\sum_{i=1}^{\infty}f_i(x)$$, term by term only when $$S_n'(x)=\sum_{i=1}^{n}f_i'(x)$$ converges uniformly. See [Rudin: Principles of Mathematical analysis(Thm:7.17) ]. So in the first step you can't differentiate as you have done. Your approach would be right if the convergence here were uniform.