# Interval of convergence using ratio test on the series $\ln(1 - x)$

I have to find the series expansion and interval of convergence for the function $\ln(1 - x)$.

For the expansion, I have gone through the process and obtained the series:

$-x - (x^2/2) - (x^3/3) - . . . - (-1)^k((-x)^k)/k$

I know that the interval of convergence will be $(-1,1)$, but am having trouble with the ratio test component to achieve this result. i.e. I am having trouble breaking down/simplifying the equation.

Thanks very much

• What is the limit that you get when trying the ratio test? – Quang Hoang Aug 19 '14 at 14:39
• I don't know, but I would think it would be x – user2768428 Aug 19 '14 at 14:40

You already know that $$\log(1-x)=-\sum_{k=1}^{\infty} \frac{x^k}{k}=\sum_{k=1}^{\infty} a_kx^k$$

Then, $$a_k=-\frac{1}{k}$$

The ratio test, then, is: $$\biggl|{\frac{a_{k+1}}{a_k}}\biggr|=\frac{\frac{1}{k+1}}{\frac{1}{k}}=\frac{k}{k+1}$$

The convergence radius $R$ is given by: $$\lim_{k\rightarrow \infty}\biggl|\frac{a_{k+1}}{a_k}\biggr|=\frac{1}{R}$$

So, $$\begin{array}{rcl} \lim_{k\rightarrow \infty}\biggl|\frac{a_{k+1}}{a_k}\biggr|&=&\lim_{k\rightarrow \infty} \frac{k}{k+1}\\ &=&1=\frac{1}{R}\\ \Rightarrow R&=&1 \end{array}$$

• Thanks for this, it does make it clearer. However I was able to get to this point but now do not know how to take it further. . . – user2768428 Aug 19 '14 at 14:46
• Also wouldn't it be x times (k/k+1)? – user2768428 Aug 19 '14 at 14:51
• @user2768428, yes i'm correcting it. – cjferes Aug 19 '14 at 14:53
• Yes, @user2768428. It should be $$\frac{xk}{k+1}$$ – Namaste Aug 19 '14 at 14:53
• @user2768428 now I'm completing the answer. – cjferes Aug 19 '14 at 14:54

Don't let the $(-1)^k$ or $(-x)^k = (-1)^kx^k$ trouble you. They have the effect of canceling each other out for odd $k$, and besides, for the ratio test, we apply it taking the absolute value of the general term $|a_k|$.

$$|a_k| = \frac{(x)^k }{k}$$

$$\frac{a_{k+1}}{a_k} = \frac{\frac{(x)^{k+1}}{k+1}}{\frac{(x)^k }{k}} = \frac{xk}{k+1}$$