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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.

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    $$\sum_{i=1}^{\infty} (-1)^{i-1}x^{i-1}=\frac{1}{1+x}$$ is valid if $|x|<1$ (convergence radius $=1$). – 2017-01-25
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    $f'(1)$ is not defined. Remember the definition of the value of a series as the limit of the partial sums. – 2017-01-25
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    Because of that "i" in the denominator, this is **not** a "geometric series". – 2017-01-25

2 Answers 2

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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.

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    Thanks for your answer.How did you find out the sequence of partial sums.I thought adding adjacent numbers in the sequence yields up zero. – 2017-01-25
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    @user408949 The sequence of partial sums is defined as $s_n=a_0+a_1+\cdots +a_n$. Or recursively $s_0=a_0$ and $s_n=s_{n-1}+a_n$. As you can see $s_0=1$, $s_1=1-1=0$, $s_2=1-1+1=1$ and so on. – 2017-01-25
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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.