0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ What is the limit that you get when trying the ratio test? $\endgroup$ – Quang Hoang Aug 19 '14 at 14:39
  • $\begingroup$ I don't know, but I would think it would be x $\endgroup$ – user2768428 Aug 19 '14 at 14:40
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. . . $\endgroup$ – user2768428 Aug 19 '14 at 14:46
  • $\begingroup$ Also wouldn't it be x times (k/k+1)? $\endgroup$ – user2768428 Aug 19 '14 at 14:51
  • $\begingroup$ @user2768428, yes i'm correcting it. $\endgroup$ – cjferes Aug 19 '14 at 14:53
  • $\begingroup$ Yes, @user2768428. It should be $$\frac{xk}{k+1}$$ $\endgroup$ – Namaste Aug 19 '14 at 14:53
  • $\begingroup$ @user2768428 now I'm completing the answer. $\endgroup$ – cjferes Aug 19 '14 at 14:54
3
$\begingroup$

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 $|a_k|$.

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.