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I'm stumped by $$\lim_{x \to \infty}\frac{1+3+5+\cdots+(2x-1)}{x+3} - x$$

My obvious first step was to get a lowest common denominator by $x(\frac{x+3}{x+3})$, giving $$\lim_{x \to \infty}\frac{1+3+5+\cdots+(2x-1)-x^2-3x}{x+3} $$

But from here I'm stumped, because with x tending to infinity, the $2x-1-x^2$ part of the numerator will be indeterminate, won't it? I was hoping to calculate the answer via the highest powers on both sides of the fraction, which I know you can do when the variable tends to infinity, but then I'd get an answer of $-\infty$ which is incorrect according to my solution book.

What did I miss?

In edit, thanks to those have responded so far, but I'm even more confused. Here's the solution in my answer book: $$\lim_{x \to \infty}\frac{(1+2x-1)\frac{x}{2}}{x+3} - x $$ $$\lim_{x \to \infty}\frac{x^2-(x+3)x}{x+3} = -3 $$

Does this make sense to any of you? You know your stuff, I'm willing to believe that either the question was badly worded or the answer is wrong.

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4 Answers 4

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Since $$1+3+5+\cdots+(2x-1)=\sum_{k=1}^{x}(2k-1)=2\cdot\frac{x(x+1)}{2}-x=x^2,$$ you'll have $$\lim_{x\to\infty}\frac{x^2}{x+3}-x=\lim_{x\to\infty}\frac{-3x}{x+3}=\lim_{x\to\infty}\frac{-3}{1+(3/x)}=-3.$$

(Here, note that $\lim_{x\to\infty}3/x=0$.) So, the answer in the book is correct.

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  • $\begingroup$ Believe it or not, in my syllabus, we haven't covered that sigma stuff. So no idea why they thought we could do this question. But I've definitely learnt from your answer - thanks! $\endgroup$
    – antgel
    Sep 2, 2014 at 16:48
  • $\begingroup$ @topper: You are welcome! $\endgroup$
    – mathlove
    Sep 2, 2014 at 16:49
  • $\begingroup$ @topper because you were supposed to get confused, plug it in for x=1,2,3,4,5..., see the "squares" pattern and figure it out from there. $\endgroup$
    – djechlin
    Sep 2, 2014 at 18:07
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Hint: you need to find a closed form for the numerator, because the number of summands depends on $x$.

$$1+3+5+\cdots+(2x-1)=(1+2+3+\cdots+2x)-(2+4+6+\cdots+2x)$$ $$=\frac{2x(2x+1)}{2} - 2\frac{x(x+1)}{2}$$

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Assuming $\;x\in\Bbb N\;$ (Otherwise I cannot understand the expression), we get:

$$\frac{1+3+5+\ldots+(2x-1)}{x+3}-x=\frac{x^2}{x+3}-x=-\frac{3x}{x+3}\xrightarrow[x\to\infty]{}-3$$

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    $\begingroup$ Shouldn't that be $3x$ in the numerator? $\endgroup$ Sep 2, 2014 at 16:28
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    $\begingroup$ Yes, it should be $-\frac{3x}{x+3}$. $\endgroup$ Sep 2, 2014 at 16:30
  • $\begingroup$ Yes, thanks: edited. $\endgroup$
    – Timbuc
    Sep 2, 2014 at 16:47
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Here is how you discover this result. Say you want to calculate.

$$\lim_{x \to \infty}\frac{1+3+5+\cdots+(2x-1)}{x+3} - x$$

So you have no idea how to do this. Why not start by plugging in $f(1)$, $f(2)$, or even $f(100)$ to see if it seems to be converging to something?

$$f(1) = \frac{1}{4} - 1 = -0.75$$ $$f(2) = \frac{4}{5} - 2 = -1.2$$ $$f(3) = \frac{9}{6} - 3 = -1.5$$ $$f(4) = \frac{16}{7} - 4 = -1.7$$

By now you may see a pattern in the numerators. The numerator is always $x^2$, although you probably want to try this for higher values of $x$ to see if you're convinced. So you can simplify this to

$$\lim_{x \to \infty}\frac{x^2}{x+3} - x$$

which you can probably solve. And on your homework, you might write "I've noticed the numerator sum is always $x^2$, but I don't know why this is true." That would still be good, partly complete answer.

When doing a problem that is unfamiliar to you, you absolutely are expected to try approaches like this. You're supposed to get stuck, then unstuck yourself. I'm disappointed in the fact that you were able to post this question and get answers that make it sound like you're magically supposed to know $1+3+5+\dots+(2n-1)=n^2$ already without making it clear how to discover this on a new problem on your own.

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