0

Calculating the series $\ln(1+x)$ using polynomial

Here's what I have tried

$$\begin{align*} (1+ax)\ln(1+x) &= \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}+a\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n+1}}{x^n} \\ &= x+ \sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^n}{n} + (2nd \space summation) \\ \end{align*}$$

As for the second summation, I have a question on distributing the lower index.

For example, I know that I can do this: $$a\sum_{n=2}^{\infty} (-1)^{n-1}\frac{x^n}{n-1}$$

However, when I redistribute the lower index, why does the exponent in $(-1)^{n-1}$ not get affected, for example why does that not then become $(-1)^{n-2}$ when I get the summation limit $\sum_{n=2}$ in this case? I ask because this is the steps shown in my textbook, and it doesn't provide a reason for this. Thanks in advance!

I'm trying to achieve this form: $$x + \sum_{n=2}^{\infty}(-1)^{n-1}\left(\frac{1}{2}-\frac{a}{n-1} \right)x^n.$$

We use the identity for the first summation: $$\sum_{n=1}^\infty x_n = x_1 + x_2 + \cdots + x_N + \sum_{n=N+1}^\infty x_n,$$

joe_bill.dollar
  • 243
  • “Here is what I tried…” What is the goal here? You haven’t told us what you are trying to do. – Thomas Andrews Dec 12 '21 at 22:55
  • How did $a$ become $a_1?$ Also, somehow you’ve got $(1+ax)\log(1+x)=\dots=\log(1+x).$ – Thomas Andrews Dec 12 '21 at 22:57
  • @ThomasAndrews I'm trying to calculate the series by re-arranging the summations, and then dividing the RHS by $(1+ax)$ to get $\ln(1+x).$ This is all fine to me, my only issue is getting the equation on the RHS into this form $x + \sum_{n=2}^{\infty}(-1)^{n-1}\left(\frac{1}{2}-\frac{a}{n-1} \right)x^n.$ I have thought of how to do this above in the explanation, my only issue is why $(-1)^{n-1}$ does not get changed at the exponent when I change the index of the summand but everything else does. – joe_bill.dollar Dec 12 '21 at 23:03
  • Your edit has $\frac{x^{n+1}}{x^n}$ where I think you mean $\frac{x^{n+1}}{n}.$ – Thomas Andrews Dec 12 '21 at 23:18
  • Can you give me another example of such a way to “calculate” a power series? I still have no idea what you mean by that. – Thomas Andrews Dec 12 '21 at 23:19
  • You mention a textbook. How about you tell us what the original problem in the textbook is, in the textbook's own words and notation? (Don't forget to cite the textbook and page number.) – anon Dec 12 '21 at 23:30
  • I think I understand what you are asking though it is being asked rather poorly. You are taking your "second summation" (it has a typo where you first introduced it btw) and you want to start the sum from the $n=2$ instead of $n=1$ so that you can combine the two sums? – Dionel Jaime Dec 12 '21 at 23:57
  • Also the $(-1)^{n-1}$ does get changed to $-1^{n-2}$. Not sure why you think it doesn't. We do have that $-1^{n-2} = - (-1)^{n-1} $. This may help you. – Dionel Jaime Dec 13 '21 at 00:12

0 Answers0