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....
 A: Cauchy theorem does not necessarily require positive terms. Further the second problem does not seem amenable to the use of Cauchy theorem. Better express it as a Riemann sum $n^{-2}\sum_{i=1}^{n}(1+(i/n))^{-2}$. Now $n$ times the above sum tends to $\int_{0}^{1}(1+x)^{-2}\,dx=1/2$ and hence desired limit is $0$.
A: 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.
A: You need to view $n$ as a constant. And that you are taking the average of $n$ things which are itself expressions in terms of $n$. So view it as a sequence $a_{r}=\frac{n}{(n+r)^2}.$
So for example for fixed n say 100. You are taking the average of a quantity say $A=\{\text{marks obtained by student +100}\}$. This is the way you have to view it.
Now the sufficient condition for the theorem to hold is precisely $\lim_{n\to\infty}a_{n}$ converges. In this case it does. So you can apply the theorem. By which I mean the nth term of an which is a function of n converges.
It's best to view this sequence as a cauchy sequence in order to conceptualize why the theorem is actually working.
A: Consider $a_n=\frac{n}{(n+n)^2}$. Clearly $\lim \limits_{n\to\infty}{a_n}=0$. Consider the quotient $\frac{\sum_{i=1}^na_n}{n}=\frac{\sum_{i=1}^n\frac{n}{(n+i)^2}}{n}=\frac{1}{(n+1)^2}+\cdots+\frac{1}{(n+n)^2}$. By Cauchy's 1st theorem on limits, $\lim \limits_{n\to\infty}\frac{\sum_{i=1}^na_n}{n}=\lim \limits_{n\to\infty}a_n$. Substituting $\lim \limits_{n\to\infty}\frac{1}{(n+1)^2}+\cdots+\frac{1}{(n+n)^2}=\lim \limits_{n\to\infty}\frac{n}{(n+n)^2}=0$ which is the case given in your book.
