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My instructor was using this example to illustrate his point-

$$ \lim_{n\to \infty} \frac{1+2+3+...+n}{n^2}$$ The numerator is an AP and the limit can be easily calculated this way to be a finite number($0.5$ in this case).

But, if we use the sum law and separate all the $\frac{r}{n^2}$ terms, the individual limits will be $0$ and so will be the answer. This is what my instructor told me to avoid.

My question is-

  1. Why can we not do this?
  2. Should we also not do this if the individual limits are finite values?
  3. Is this specific to the sum law?
Robin
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  • One of the reasons is that every time you have more and more terms/summands in the numerator which you can't control. In such cases we use either the squeeze theorem or, e. g., Stolz-Cesaro. – PinkyWay Jun 18 '20 at 14:51
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    $$\frac{1+1+\cdots+1}n$$ with $n$ ones on the top illustrates the same point in an even simpler way. – Angina Seng Jun 18 '20 at 14:53
  • See tha answer by Paramanand Singh. – PinkyWay Jun 18 '20 at 14:56

1 Answers1

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The sum law, or any of the other limit laws, is usually stated only for the case of 2 terms. By induction, we may extend it to 3, 4, 5, or any other fixed finite number of terms, where fixed means that the number does not depend on the limit variable $n$. However, as your example shows, we may not extend it to the case where the number of terms depends on $n$.

Ted
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  • Yes, we can. We can also extend it to any fixed, finite combination of sums, products, etc. – Ted Jun 18 '20 at 15:47