<s>[INST] cauchy's first theorem on limits of sequences Cauchy's first theorem on limits goes like this
If$\ <f_n> $ be a sequence of positive terms and $$ \lim_{\ n\to\infty}\ f_n=l$$ Then
$$ \lim_{ n\to\infty}\ [\ \frac{f_1+f_2+\dots+f_n}{n}]=l$$
Now this is an example of its application. 
$Q)$ Find the value of $$ \lim_{\ n \to \infty}\frac{1}{n}[1+\frac{1}{2}+\frac{1}{3}\dots+\frac{1}{n}] $$
$A)$ By cauchy's theorem if $$ f_n=\frac{1}{n} \ and \lim_{\ n\to\infty}\frac{1}{n}=0 $$ Then $$ \lim_{\ n\to\infty}\frac{1}{n}[1+\frac{1}{2}+\frac{1}{3}\dots+\frac{1}{n}]=0$$ I understand this example. Now here is another example and its solution which I found in various texts but I don't understand how can we apply cauchy's theorem to it
$Q)$ Find the value of  $$ \lim_{\ n\to\infty}[\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{2n^2}] $$
$A)$ Multiply and divide by n 
 $$ \lim_{\ n\to\infty}\frac{1}{n}[\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\dots+\frac{n}{2n^2}] $$
Let $$ <f_n>=\frac{n}{(n+n)^2}=\frac{n}{4n^2}$$
Then, $$ \lim_{\ n\to\infty}f_n=\lim_{n\to\infty}\frac{1}{4n}=0$$
 By cauchy's theorem
 $$ \lim_{\ n\to\infty}[\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{2n^2}]=0 $$
As you can see there is clearly a distinction between the first and second question
In th first we have $ 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$
while in the second one we have $\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\dots+\frac{n}{(2n)^2} $
I see how the first one can be written as the sum of sequence $\frac{1}{n}$
But how can the second one written as the sum of the sequence $\frac{n}{(2n)^2}$
Please help....
 [/INST] Answer to the second question.
Observe that for all $n \ge k$,
$$ \frac{1}{2n} \le \frac{1}{n+k} \le \frac{1}{n}$$
Thus,
$$\left( {\frac{1}{{{{(2n)}^2}}} + \frac{1}{{{{(2n)}^2}}} +  \ldots  + n{\mkern 1mu} {\mkern 1mu} terms} \right) \le \left( {\frac{1}{{{{(n + 1)}^2}}} + \frac{1}{{{{(n + 2)}^2}}} +  \ldots  + \frac{1}{{2{n^2}}}} \right) \le \left( {\frac{1}{{{n^2}}} + \frac{1}{{{n^2}}} +  \ldots  + n{\mkern 1mu} {\mkern 1mu} terms} \right)$$
$$\implies \frac{1}{4n} \le \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{2n^2} \le \frac{1}{n}$$
The answer follows by the Sandwich theorem.
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