The first part of my question (for my homework) states "Find the Taylor series about $x=0$ of the function $f(x)=\frac{1}{(1-x)^2}$ "...

I have found the taylor series of $\frac{1}{(1-x)^2}$ to be $$\sum_{n=0}^\infty\dfrac{\dfrac{1}{(1-a)^2}^n{(x-a)^n}}{n!}$$ and the Maclaurin series to be $$\sum_{n=0}^\infty(n+1)x^n$$

I have a couple of questions...

  1. Am I correct in starting my bound as $n=0$? I wasn't sure because some examples have $n=1$. Maybe that's just if the function is undefined for $n=0$?
  2. I have found both the taylor series and the maclaurin series as the question asks for the "Taylor series" about $x=0$ which i thought was specifically called a Maclaurin series? But they are different series, so... I am asked to then find the radius and interval of convergence of the answer but I am really confused as to whether I am meant to use the Taylor series version or the Maclaurin series version to do this. Can someone clarify which series out of the ones I found to use for this part?
  3. The last half of the question states "Hence or otherwise determine the sum of the series summed from $n=0$ to infinity of $\frac{(n+1)}{2^n}$... Does that part have anything to do with either of the series I found for the first half? Or is it just a new part of the question and I should apply the same techniques to find the sum of that series as any other and just ignore the first half of the question to complete this part?
  • $\begingroup$ Learn how to format @ math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Mohammad Zuhair Khan Oct 27 '18 at 4:06
  • $\begingroup$ Okay the editing is fine now, can anyone please help with my questions? I'm really stuck. $\endgroup$ – User1997 Oct 27 '18 at 4:34
  • $\begingroup$ Your "Taylor series" adds to $$\frac{\exp(x-a)}{(1-a)^2}$$ so cannot be correct. In any case a Taylor series about $0$ is a Maclaurin series. $\endgroup$ – Angina Seng Oct 27 '18 at 4:38
  • $\begingroup$ @LordSharktheUnknown How does it add to that? $\endgroup$ – User1997 Oct 27 '18 at 4:44
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    $\begingroup$ The Taylor series at $0$ is the Maclaurin series, which you have found. All that stuff you're writing with "general" $a$ is irrelevant (and wrong) and should just be deleted. The power series representation equals the function where the power series converges. This stuff all has real numerical meaning. Look at the graphs of $y = 1/(1-x)^2$ together with the graphs of the partial sums of the Taylor series at $0$: $y = 1$, $y = 1 + 2x$, $y = 1 + 2x + 3x^2$, and so on. Around $x = 0$ they should look quite close to each other. $\endgroup$ – KCd Oct 27 '18 at 5:26

Your last question: on your second attempt you successfully computed the Maclaurin series as $$\frac1{(1-x)^2}=\sum_{n=0}^\infty(n+1)x^n.\tag{$*$}$$ You are enjoined to calculate $$\sum_{n=0}^\infty\frac{n+1}{2^n}.$$ This looks very similar to the RHS of $(*)$. Can you obtain it from the RHS of $(*)$ by making a substitution?

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  • $\begingroup$ Do you mean that i substitute something into $$(*)$$ to make it like the other series? x would have to equal 2 and n would have to be -n $\endgroup$ – User1997 Oct 27 '18 at 5:17
  • $\begingroup$ @KaylaMartin Very big hint: $(\frac{1}{a})^n = \frac 1{a^n}$. $\endgroup$ – Deepak Oct 27 '18 at 5:18
  • $\begingroup$ Yeah i can see thats what i need to do but i cant really tell how to do that by substition? What am i allowed to do to it? Can i let $$x^n$$ = $$\frac{1}{2^{n}}$$ ? Thats all i can think of. $\endgroup$ – User1997 Oct 27 '18 at 5:20
  • $\begingroup$ @KaylaMartin What happens when $x= \frac 12$? $\endgroup$ – Deepak Oct 28 '18 at 3:16

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