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I have a doubt on applying Cauchy's 1st theorem on limits. Kindly enlighten me....

Example 1: $\lim_{n\to \infty} \frac {1}{n} (1+\frac {1}{2}+....\frac{1}{n})$.

Here I am taking $a_n = 1/n \to 0 $ as $n \to \infty$. So $ \frac{a_1+a_2+...+a_n}{n} \to 0\implies$ given example 1 limit converges to 0

Now take example 2:

Example 2: $ \lim_{n\to \infty} \frac {1}{n} (\frac{1}{n+1}+\frac {1}{n+2}+....\frac{1}{n+n})$

Here in textbook I have seen

$a_n = 1/(n+n) \to 0$ and so $\lim_{n\to \infty} \frac {1}{n} (\frac{1}{n+1}+\frac {1}{n+2}+....\frac{1}{n+n}) \to 0$ ------>(1)

If $a_n = 1/(n+n) = 1/(2n)$

can we take in this way? $\lim_{n\to \infty} \frac {1}{n} (\frac{1}{2}+\frac {1}{2*2}+....\frac{1}{2*n})$ ------>(2)

Is (2) wrong? If so, can you pls elaborate?

Magneto
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  • Can you tell us which textbooks are you using? For now the argument seem very sloppy (though $(1)$ is correct). –  Jun 15 '18 at 17:52
  • (2) i have written. (1) is present in any text book or google. I am confused with 2 n's which only 1 n will be substituted and other n is kept as it is – Magneto Jun 15 '18 at 17:55
  • I am saying that if you are using Cauchy's first theorem with $a_n = \frac{1}{n+n}$, then you should end up in (2) instead of (1). If a textbook claims that (1) is true by Cauchy's first theorem, then the argument is very sloppy. –  Jun 15 '18 at 17:59
  • Sir pls see example (2) in https://math.stackexchange.com/questions/1930373/limits-and-cauchys-first-theorem – Magneto Jun 15 '18 at 18:05
  • The argument by the questioner there is incorrect..... –  Jun 15 '18 at 18:07
  • Sir i have seen examples(like example (2) in the link) under cauchy 1st theorem on limits where there will be two n's. 1 n will be kept as fixed as other n is moved from 1 to n. Pls kindly elaborate if possible – Magneto Jun 15 '18 at 18:15
  • @Magento Have you got the answer to this problem ? I have encountered this myself in numerous occasions in textbooks and can't understand the reason. If you are through with this problem, request you to kindly explain this. – llecxe Nov 26 '20 at 22:14

1 Answers1

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You can not apply the theorem for the limit given in example 2 (because each element in the sequence $\left\{\frac{1}{1+n},\ldots,\frac{1}{n+n} \right\}$ is dependent on $n$). But you can calculate the limit as follows

$$ 0 \le \lim_{n\to\infty} \frac1n \, \left( \frac{1}{1+n} + \ldots+ \frac{1}{n+n} \right) \le \lim_{n\to\infty} \frac1n \, \left( \frac{n}{n+1} \right) \to 0 $$ Thus, by the squeeze theorem $\lim_{n\to\infty} \frac1n \, \left( \frac{1}{1+n} + \ldots+ \frac{1}{n+n} \right)=0$.

yakobyd
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