Question on the sum $\sum_{n=1}^{\infty}\frac{x^n}{n} = -\ln(1-x)$

Question

$f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$.

$f'(x) = \displaystyle\sum_{n=1}^{\infty}x^{n-1} = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$

If $f'(x) = \displaystyle\frac{1}{1-x}$, then $f(x) = \displaystyle\int\frac{1}{1-x}dx$

But $\displaystyle\int \frac{1}{1-x}dx = -\ln(1-x) + C$, and so $f(x) = -\ln(1-x) + C$

Do we need a definite integral here so there will be no $C$ and will agree with how $f(x)$ was originally defined? If so what would the definite integral be? Or maybe there is no definite integral at all?

Answer 1

You're right: $f'(x)=\frac1{1-x}$, so $f(x)$ must be $-\ln(1-x)+C$ for some $C$. Now, to find what $C$ is, try letting $x$ be $0$ and solving for $C$.

Or, if you want to do a definite integral, start with: $$1+t+t^2+\dotsb=\frac1{1-t}$$ Take the definite integral from $0$ to $x$: \begin{align} \int_0^x(1+t+t^2+\dotsb)\operatorname d\!t&=\int_0^x\frac1{1-t}\operatorname d\!t\\ x+\frac{x^2}2+\frac{x^3}3+\dotsb&=-\ln(1-x) \end{align}

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Note that if we take $x=-1$, we get the famous series for $\ln2$: $$\ln2=1-\frac12+\frac13-\dotsb$$ which can be proven without calculus if you try hard enough.

Answer 2

There are various things going on here. One is the issue of the $+C.$ What you wrote might better be put

If $f'(x) = \displaystyle\frac{1}{1-x}$, then by definition of the indefinite integral $\displaystyle\int\frac{1}{1-x}dx=f(x)+C$

Also , by definition of definite integrals

If $f'(x) = \displaystyle\frac{1}{1-x}$, on $[a,b]$ then $ \displaystyle\int_a^b\frac{1}{1-t}dt=f(b)-f(a)$

Note that an arbitrarily chosen constant of integration $C$ will cancel out.

I changed the variable of integration so that I could say that for $|x| \lt 1,$ $$\int_0^x\frac1{1-t}dt=-\ln(1-t)\mid_0^x=\left(-\ln(1-x)\right)-(-\ln(1-0))=\ln\left(\frac1{1-x}\right)-0$$

There is also the matter of term by term integration and differentiation of power series. Notice that this is one of the interesting cases where

$f(x) = = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$ and also $x=-1$ but not $x=1.$

$f'(x) = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$ but not for $x= -1$ or $x=1.$