Question

$$S_\infty=\sum^\infty_{n=1}\frac{(-1)^n}{n}$$

• Determine whether \(S_\infty\) converges, and what is the converged value (if it does).

Solution

This is a bit of a magical substitution.

Notice the Taylor series expansion of \(\ln\) around \(x=a\).

$$\ln(a+x)=\ln a+\frac{0!}{1!\cdot a}x-\frac{1!}{2!\cdot a^2}x^2+\frac{2!}{3!\cdot a^3}x^3-\dots$$

Substitute \(a=1\)

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$$

Substitude \(x=1\)

$$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$$

Compare it to our original sum

$$S_\infty=-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\dots$$

We can see that \(S_\infty=-\ln 2\)

Extension

Referencing back to EyeBeam’s post on 31st May, you can also proof that the harmonic series \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots\) diverges. Which is left as an excercise for the reader.