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$$ \frac{1 + 1 + 1 + \cdots}{1 + 2 + 3 + \cdots} = \lim_{n \to \infty} \frac{1}{(n+1)/2} = 0 $$

If $n$ goes to infinity, we can image that a bit by taking a very big number.

Like $1.000.000.000$

$1.000.000.000+1=1.000.000.001$

$1.000.000.001/2=500.000.000,5$

Now how does the $1 + 2 + 3 + ...$ make sense?

Jeel Shah
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user8005
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3 Answers3

4

They're using that $$1+2+3+\dots+n=\frac{n(n+1)}2$$

For example $1+2+3=6=\dfrac{3\cdot 4}{2}$

On the other hand $$\underbrace{1+1+1+\cdots+1}_{n \;\;\rm times}=n$$

Thus $$\frac{1+1+1+\cdots}{1+2+3+\cdots}=\frac{n}{\dfrac{n(n+1)}2}=\frac{2}{n+1}$$

Of course, we're interpreting $$\frac{1+1+1+\cdots}{1+2+3+\cdots}$$ as $$\lim\limits_{n\to\infty}\frac{\sum_{i=1}^n 1}{\sum_{i=1}^n i}$$

If we had, for example $$\lim\limits_{n\to\infty}\frac{\sum_{i=1}^{n^2} 1}{\sum_{i=1}^n i}$$

then we would say that $$\frac{1+1+1+\cdots}{1+2+3+\cdots}=2$$

We can write this more suggestively as $$\frac{{1 + (1 + 1 + 1) + (1 + 1 + 1 + 1 + 1) + \cdots }}{{1 + 2 + 3 + \cdots }} = 2$$ to make clear how many terms we add in each step.

The moral of the story is that you cannot have "voids" in your notation. Writing something like $$\frac{1+1+1+\dots}{1+2+3+\dots}=\text{something}$$

should be better replaced by something informative and clear that really says what is going on.

Pedro
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  • please elaborate on: $$\frac{{1 + (1 + 1 + 1) + (1 + 1 + 1 + 1 + 1) + \cdots }}{{1 + 2 + 3 + \cdots }} = 2$$ I don't understand the grouping : (1+3+5)/(1+2+3) – user8005 May 20 '13 at 00:54
  • That's because $$\sum_{i=1}^{n^2} 1=n^2$$ – Pedro May 20 '13 at 00:57
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Look user, you are brining up a great issue here! The numerical expression itself is pretty meaningless if we do not know the context, i.e. the formula where this expression came from. With regrouping symbols I can get virtually anything out of the given numerical expression, because, as stated, it is an infinity / infinity situation, which is indeterminate. The word "indeterminate" is really appropriate here. Sure, the answer could be 1/2 if the problem arises from n/2n, but with grouping symbols, the problem also could arive from 2n/2n in which the answer would be 1. If I group enough 1's together, I can also get 4 as an answer. As long as we do not know from which expression your numerical problem comes from, there is no point arguing what answer it has to be.

imranfat
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It doesn't in general make sense to take ratios of infinite series because convergence may not be assured. In this case, the numerator and denominator diverge so you have to be very careful about what you mean by

$$\frac{1+1+1+\cdots}{1+2+3+\cdots}.$$

The best way to interpret this ratio is as the following:

$$\frac{1+1+1+\cdots}{1+2+3+\cdots} = \lim_{n\rightarrow\infty}\frac{\sum_{i=1}^n1}{\sum_{i=1}^ni} = \lim_{n\rightarrow\infty} \frac{n}{\frac{n(n+1)}{2}} = \lim_{n\rightarrow\infty}\frac{2}{n+1} = 0.$$

Cameron Williams
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  • I strongly disagree that this is "the best way to interpret this ratio". Filip's point in bringing up that ratio in the first place was that that is not a good way to interpret it. – MJD May 20 '13 at 01:24
  • It's the way OP was thinking about it and if someone wrote that, I would assume they meant what I wrote because often people are very loose with mathematical notation. Admittedly I did not see the post by Filip though. I just assumed OP meant what I was addressing. But of course, that ratio is terribly ill-defined. – Cameron Williams May 20 '13 at 05:15