Squeeze theorem Is there any trick when it comes two find two sequences with the same limit to prove that a third sequence that is between them has also a limit? There are a whole bunch of sequences with the same limit.
As an example, $\lim_{n \to \infty}\{\frac{n}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\}$.
 A: You can see the correct estimate
$a_n\leq n^2/(2n^2+1)$ since you have a sum of $n$ things less than or equal to $n/(2n^2+1)$.
$n^2/(2n^2+n) \leq a_n$ since you have a sum of $n$ things greater than or equal to $n/(2n^2+n)$.
In general, you try to reduce the complexity of the expression for the $n$th term of a sequence and see how the original sequence compares.  It often helps to know that part of the expression is bounded.  For instance, we know that $-1\leq$sin$2n\leq 1$, so we have 
$-1/(1+\sqrt n)\leq ($sin$2n)/(1+\sqrt n)\leq 1/(1+\sqrt n)$, 
whence by the squeeze theorem
lim$_{n\to\infty} ($sin$2n)/(1+\sqrt n)=0$.
A: In this case, I would just take the minimum term and the maximum term and replace all terms by that term. That is, since there are $n$ terms in the sum,
$$
n\frac{n}{2n^2+n}\le\frac{n}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\le n\frac{n}{2n^2+1}
$$
Then we can simplify the left and right sides to be things whose limit can more easily be seen:
$$
\frac1{2+1/n}\le\frac{n}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\le \frac1{2+1/n^2}
$$
