Showing a series converges to a function

Question

I have calculated the Taylor series for $f(x)=\ln(1+x)$ centered at $x=0$ with some help and have gotten the following,

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^n$$,

I then found the values of $x$ for which the series converges, through the ratio test I found these values to be $-1<x<=1$. I am now trying to prove that the series converges to the function $ln(1+x)$ for these values of $x$ that I have found.

So how would I prove that it converges to the function?

Answer 1

I thought it might be instructive to show convergence using only the sum of a geometric progression and an elementary inequality. To that end we proceed.

Note that we can write

$$\begin{align} \sum_{n=1}^N\frac{(-1)^{n-1}x^n}{n}&=\sum_{n=1}^N (-1)^{n+1}\int_0^xt^{n-1}\,dt\\\\ &=\int_0^x \sum_{n=1}^N (-t)^{n-1}\,dt\\\\ &=\int_0^x\frac{1-(-t)^N}{1+t}\,dt\\\\ &=\log(1+x)-(-1)^N\int_0^x\frac{t^N}{1+t}\,dt \end{align}$$

Now, note that for $|x|<1$

$$\begin{align} \left| \int_0^x\frac{t^N}{1+t}\,dt\right|&\le \frac{1}{1-|x|}\int_0^x t^n\,dt\\\\ &=\frac{x^{N+1}}{(1-|x|)(N+1)}\\\\ &\to 0\,\,\text{as}\,\,n\to \infty \end{align}$$

Therefore, for $|x|<1$, we have

$$\lim_{N\to \infty}\sum_{n=1}^N\frac{(-1)^{n-1}x^n}{n}=\log(1+x)$$

as was to be shown!

Answer 2

According to Lagrange remainders, we have

$$P_k(x)=\sum_{n=1}^k\frac{(-1)^{n+1}}nx^n$$

For some $\xi\in(-1,1)$, we then have

$$\ln(1+x)-P_k(x)=\frac{\xi^{k+1}}{k+1}$$

But since $\lim_{k\to\infty}\frac{\xi^{k+1}}{k+1}=0\forall\xi\in(-1,1)$, it follows that

$$\ln(1+x)=\lim_{k\to\infty}P_k(x)$$

At $x=1$, we can show convergence through the partial sums:

$$\begin{align}\sum_{n=1}^k\frac{(-1)^{n+1}}n&=\sum_{n=1}^k(-1)^{n+1}\int_0^1t^{n-1}\ dt\\&=\int_0^1\sum_{n=1}^k(-1)^{n+1}t^{n-1}\ dt\\\text{(geometric series)}&=\int_0^1\frac{1-(-t)^k}{1+t}\ dt\\&=\int_0^1\frac1{1+t}\ dt+(-1)^{k+1}\int_0^1\frac{t^k}{1+t}\ dt\\&=\ln(2)+(-1)^{k+1}\int_0^1\frac{t^k}{1+t}\ dt\end{align}$$

We can then squeeze the last integral, since:

$$0<\int_0^1\frac{t^k}{1+t}\ dt<\int_0^1t^k\ dt=\frac1{k+1}\stackrel{k\to\infty}\longrightarrow0$$

Answer 3

Such series is the Taylor series of $\log(1+x)$ at the origin. The Taylor series is unique. The radius of convergence of the power series is one. $\log(1+x)$ is an analytic function in a neighbourhood of the origin. It follows that $$ \log(1+x) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}x^n \tag{1}$$ holds for any $x\in(-1,1)$. As an alternative, for any $z\in(-1,1)$ we have $$ \frac{1}{1+z} = \sum_{n\geq 0} (-1)^n z^n\tag{2} $$ hence by integrating both sides of $(2)$ over the interval $(0,x)$ we get $(1)$ for any $x\in(-1,1)$. We are allowed to exchange $\int$ and $\sum$ due to the absolute convergence of the RHS of $(2)$.