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I am able to evaluate the limit $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$ for a given $n$ using l'Hôspital's (Bernoulli's) rule.

The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:

  • Substitute $y = x - 1$. This leads nowhere, I think.
  • Find the Taylor polynomial. Makes no sense, it is a polynomial.
  • Divide by major term. Dividing by $x$ got me nowhere.
  • Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$.
  • Simplify the expression. I do not see how I could.
  • Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it.

If somebody sees a simple way, please do let me know.


Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:

  • $\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ where $m, n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are given, or
  • $\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}$.

It sems that all limits in the form $\lim_{x \to a} \frac{f(x)}{x - a}$ where $a \in \mathbb{R}$, $f(a) = 0$ and for which $\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.

This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :).

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8 Answers 8

up vote 87 down vote accepted

Let $$f(x)=x+x^2+\cdots+x^n-n$$ then by the definition of the derivative we have

$$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1}= \lim_{x \to 1}\frac{f(x)-f(1)}{x - 1}=f'(1)\\=[1+2x+3x^2+\cdots+nx^{n-1}]_{x=1}= \frac{n(n + 1)}{2}$$

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1  
It means that we should substitute $x$ by $1$. –  Sami Ben Romdhane Jul 25 '13 at 12:20
10  
Then this is truly an ingenious solution. –  David Čepelík Jul 25 '13 at 12:30
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So the original is just the integral of a function summing the positive natural numbers? Pauling would be proud. –  Kyle Hale Jul 25 '13 at 15:29
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I almost cried when I saw this solution. –  Andrew Larsson Jul 25 '13 at 17:44
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@RossMillikan shouldn't it not matter? if $f(x)$ contains the $-n$ then $f(1) = 0$, and if $f(x)$ does not, $f(1) = n$, but either way $f(x) - f(1)$ doesn't change and since $n$ is a constant, $f'(x)$ looks the same. –  Eugene Bulkin Jul 25 '13 at 23:22

Well, you can use that \begin{align} x-1&=\left(x-1\right)\cdot 1,\\ x^2-1&=\left(x-1\right)\cdot\left(x+1\right),\\ x^3-1&=\left(x-1\right)\cdot(x^2+x+1),\\ \ldots\\ x^n-1&=\left(x-1\right)\cdot(x^{n-1}+\ldots+1),\\ \end{align} Now sum the left hand sides and the right hand sides, divide by $x-1$ and consider the limit $x\rightarrow 1$.

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Hints:

$$x+x^2+\ldots+x^n-n=(x-1)+(x^2-1)+\ldots (x^n-1)=(x-1)\left(1+(x+1)+\ldots\right)$$

Also

$$1+(x+1)+(x^2+x+1)+\ldots+(x^{n-1}+\ldots+x+1)\xrightarrow [x\to 1]{}1+2+\ldots+n=\frac{n(n+1)}2$$

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The given limit is $$\lim_{x\rightarrow 1}\frac{\sum_{k=1}^nx^k-n}{x-1}\\ =\lim_{x\rightarrow 1}\frac{\sum_{k=1}^n(x^k-1)}{x-1}\\ =\sum_{k=1}^n \lim_{x\rightarrow 1} \frac{(x^k-1)}{x-1}$$

Now, $$\lim_{x\rightarrow 1} \frac{(x^k-1)}{x-1}\\ =\lim_{x\rightarrow 1} (\sum_{j=0}^{k-1}x^j)=k$$

Hence the given limit becomes $$\sum_{k=1}^n k=\frac{n(n+1)}{2}$$

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You can use induction:

$$\frac{x + x^2 + \dots + x^n + x^{n+1} - (n+1)}{x - 1} =\\ \frac{x + x^2 + \dots + x^n - n}{x - 1} +\frac{x^{n+1}-1}{x-1}=\\ \frac{x + x^2 + \dots + x^n - n}{x - 1} +(1+x+x^2+\ldots+x^n)\xrightarrow[x\to 1]{} \frac{n(n + 1)}{2}+(n+1)=\frac{(n+1)(n + 2)}{2}.$$

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I am a little lost. How is that you are getting different result? Or is $\frac{(n + 1)(n + 2)}{2}$ only a part of the solution? –  David Čepelík Jul 25 '13 at 12:28
    
@DavidČepelík No, $\frac{(n+1)(n+2)}{2}$ is the limit for $\frac{x+x^2+\ldots+x^n\color{brown}{+x^{n+1}}-n\color{brown}{-1}}{x-1}$ and not for $\frac{x+x^2+\ldots+x^n-n}{x-1}$. –  P.. Jul 25 '13 at 12:32
    
Of course, that is the induction. Thank you! –  David Čepelík Jul 25 '13 at 12:37
    
So for $n = 1$ and $x \to 1$, we get $a_1 = \lim_{x \to 1} \frac{x - 1}{x - 1} = 1$ and $a_{n + 1} = a_n + (n + 1)$. The closed form for $a_n$ is then $\frac{n(n + 1)}{2}$. Correct? –  David Čepelík Jul 25 '13 at 12:42
    
@DavidČepelík: Yes that's right! –  P.. Jul 25 '13 at 13:08

$$\frac{x + x^2 + \dots + x^n - n}{x - 1} =\frac{1+x + x^2 + \dots + x^n - n-1}{x - 1} =\frac{\frac{x^{n+1}-1}{x-1}-(n+1)}{x-1}$$

Putting $x-1=y,$

$$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \lim_{y\to0}\frac{\frac{(1+y)^{n+1}-1}y-(n+1)}y$$

$$=\lim_{y\to0}\frac{\frac{1+\binom {n+1}1y+\binom {n+1}2y^2+ O(y^3)-1}y-(n+1)}y\text{ (using Binomial Expansion)}$$

$$=\lim_{y\to0}\frac{(n+1)+\frac{n(n+1)}2y+ O(y^2)-(n+1)}y$$

$$=\frac{n(n+1)}2\text{ as }x\to1,y\to0\implies y\ne0$$

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If you already know l'Hopital's rule, Sami Ben Romdhane's answer isn't telling you anything you don't already know. When you have $$\lim_{x\to a}\frac{f(x)}{x - a}$$ and $f(a) = 0$ (otherwise, the limit is infinite on either side of $a$), then l'Hopital's rule says that the limit is the same as $$\lim_{x \to a}f'(x)= f'(a).$$ The only hard part then, in this case, is evaluating $f'(a)$. Here, $f(x) = x + x^2 + \cdots + x^n - n$, so $f'(x) = 1 + 2x + \cdots + nx^{n - 1}$, so $f'(1) = 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}$ by a famous identity.

In fact, you can think of the definition of the derivative as just a special case of l'Hopital's rule (it isn't really, because l'Hopital's rule depends on the definition of the derivative, not the other way around, but it's useful to think of it this way).

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I do not think this is how he approached the problem; my guess is he noticed that the function and the derivative of the numerator are bound together by the definition of derivative of a function. I did not realize that he used the same principle Bernoulli's rule is built upon, thanks for pointing it out. –  David Čepelík Jul 25 '13 at 16:23

Your objection to using l'Hospital's rule is on the basis that it feels like it's "too powerful" a tool for the problem, right?

Lets go all the way to the other end, then, and prove the limit using just the definition.

$$\lim_{x\to a} f(x) = L \iff \forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ni \left| x-a \right| \le \delta \Rightarrow \left| f(x)-L \right| \le \varepsilon$$

Now, all we need to do is figure out a way that we can always pick a $\delta$ small enough to keep the function within $\varepsilon$ of $L$.

The particular limit in question: $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$

From that, we can take: $$\begin{align*} a&=1\\ f(x)&=\frac{x + x^2 + \dots + x^n - n}{x - 1}\\ L&=\frac{n(n + 1)}{2} \end{align*}$$

So we need to solve $$\left| \frac{x + x^2 + \dots + x^n - n}{x - 1} - \frac{n(n + 1)}{2}\right| < \varepsilon$$ for $\left|x-1\right|$.

I really don't feel like doing any of that crunchwork solving that equation. If someone out there does feel like it, please do so, and edit it into my answer. For now, though, I am going to skip ahead bunch of steps, assume we solved it, and have our solution of $$\left|x-1\right| \ge g(\varepsilon)$$

Now, we know from the definition that $\left| x-1 \right| \le \delta$, so we can conclude that if we pick $\delta = g(\varepsilon)$, we ensure that the value of $f(x)$ is within $\varepsilon$ of $L$, satisfying our definition of the limit, and proving that $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$

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Thanks for your answer. You already know the limit exists and you know what it equals to ($\frac{n(n+1)}{2}$ was not my first guess). Finding what a limit equals to and proving the solution is correct by checking it satisfies the definition of the limit are two completely different exercises. (Similar to the difference between the P and NP classes in the theory of computational complexity.) I am aware I gave correct solution along with the limit, maybe that is why we did not understand each other. Sorry about it. –  David Čepelík Jul 26 '13 at 20:01
    
@DavidČepelík Personally, I think that if you prove the limit without using your super-powered tool, it doesn't matter if you used it to find the value of the limit, since the validity of your answer then does not depend on the validity of the method you applied. –  AJMansfield Jul 29 '13 at 12:57

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