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again! Last question for the moment... possibly. Continuing with the theme of “little-o” limits, I present the following question:

Find the limit of:

$\lim_{x\rightarrow1}\frac{x+x^{2}+...+x^{n}-n}{x-1},\,\, n\epsilon\mathbb{N}$

Now, at first I thought this question would be deceptively easy... I tried using the fact that the above has a finite series in it, so replacing it with the finite series should simplify it. This did not help, however.

My finite summation came to (after a change in variable to let $x=t+1$):

$\frac{(t+1)-(t+1)^{n+1}}{-t}$

Using the fact that $(1+x)^{n}\simeq1+nx+o(x)$:

$\frac{t+1-[1+(n+1)t+o(x)]}{-t}=\frac{t-t(n+1)+o(x)}{-t}$

Substituting back into the original limit equation:

$\lim_{t\rightarrow0}\frac{\frac{t-t(n+1)+o(x)}{-t}-n}{t}$

Which, after simplifying down, seems to limit to anything except the actual answer of $\frac{n(n+1)}{2}$. Have I made a faulty assumption? Or is my strategy not appropriate?

EDIT: The above strategy seems to yield me exclusively $\frac{o(x)}{t^{2}}$

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  • $\begingroup$ Your work is correct. You remember me a good moment:-) $\endgroup$
    – user63181
    May 3 '14 at 10:29
  • $\begingroup$ Thanks for the link! I hadn't considered it like that. It does use a tool different to the one we're considering in this chapter though, so I was wondering if there was a way to solve it considering only series, convergence, limit properties and "little-o" order theorems? $\endgroup$
    – Yoshi
    May 3 '14 at 10:31
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Your method is nice but needs some correction:

Let $x=1-t$ so we have

$$\frac{x+x^2+\cdots+x^n-n}{x-1}=-\frac{\displaystyle\sum_{k=1}^n(1-t)^k-n}{t}\\=_0-\frac{\displaystyle\sum_{k=1}^n(1-kt+o(t))-n}{t}=_0\sum_{k=1}^nk+o(1)\xrightarrow{t\to0}\frac{n(n+1)}{2}$$

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  • $\begingroup$ Ok, so I understand the change in variable, but what exactly is happening after the point where the order is introduced? I don't see that last step. $\endgroup$
    – Yoshi
    May 3 '14 at 10:50
  • $\begingroup$ Just we simplify the expression and notice that $$o(t)+o(t)=o(t)$$ $\endgroup$
    – user63181
    May 3 '14 at 10:55
  • $\begingroup$ I'm afraid I still don't quite see. Once the expression inside the series is expanded to include the order, we just divide each member by t (Since there's a t on the bottom of the fraction)? I've also read a few of the other posts on this question, and I can never seem to understand why in the last step, that n(n+1)/2 seems to simply materialize without warning. $\endgroup$
    – Yoshi
    May 3 '14 at 11:02
  • $\begingroup$ It's well known (unless that's what it should be) that the sum of $1+2+\cdots+n=\frac{n(n+1)}{2}$ and that $o(1)$ is a function that tends to $0$ when $t\to0$. $\endgroup$
    – user63181
    May 3 '14 at 11:07
  • $\begingroup$ Ok, that's making more sense. How does n cancel out with the 1 during our simplifying? I understand that o(t) = t.o(1). $\endgroup$
    – Yoshi
    May 3 '14 at 11:14
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Your original idea to use geometric sums is also fine, you just need to apply the quadratic Taylor polynomial instead of the linear one, $(1+t)^{n+1}=1+(n+1)t+\binom{n+1}2t^2+O(t^3)$ \begin{align} \frac{x+x^2+...-x^n-n}{x-1}&=\frac{x^{n+1}-x-n(x-1)}{(x-1)^2}=\frac{(1+t)^{n+1}-(1+(n+1)t)}{t^2}\\ &=\tbinom{n+1}{2}+\tbinom{n+1}{3}t+\tbinom{n+1}{4}t^2+...+t^{n-2} \end{align}

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