0

Background

In a previous problem, I have shown that the Taylor polynomial around $x=0$ for $f(x) = \ln(1-x)$ is $$T_nf(x) = -\sum_{k=1}^{n}\frac{x^k}{k}$$

Problem

Use this Taylor expansion to show that $\displaystyle \ln2 = \sum\limits_{k=1}^{\infty}\frac{1}{n2^n}$.

My thoughts

I figured, hey, $\ln(2) = \ln(1-(-1))$ so with $x=(-1)$ I can just plug it into that Taylor expansion. But apparently that Taylor polynomial is only valid for $x\in(-1, 1)$, so I can't do that.

So that's where I'm stuck.

Any help appreciated!

Alec
  • 4,094

1 Answers1

1

The Taylor polynomial is actually valid at $x=-1$, which is right on the boundary. Using it, we have

$$\ln(2)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$$

But this isn't supposed to be the answer! Using log rules, however, one may note that

$$\ln(2)=-\ln(1/2)=-\ln(1-1/2)=\sum_{n=1}^\infty\frac1{n2^n}$$

If you hadn't noticed this, however, you could apply an Euler transform to get

$$\ln(2)=\sum_{n=0}^\infty\frac{(-1)^n}{n+1}=\sum_{n=0}^\infty\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{k+1}$$

Prove that $\sum_{k=0}^n\binom nk\frac{(-1)^k}{k+1}=\frac1{(n+1)}$ so that we have

$$\ln(2)=\sum_{n=0}^\infty\frac1{(n+1)2^{n+1}}=\sum_{n=1}^\infty\frac1{n2^n}$$

Simply Beautiful Art
  • 74,685