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?
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 $\lim_{n\to\infty} \frac1n \, \left( \frac{1}{1+n} + \ldots+ \frac{1}{n+n} \right)=0$.