# 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, $\lim_{n \to \infty}\{\frac{n}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\}$.

• Allsony, Can you give any example question? – Saharsh Jun 26 '14 at 16:22
• $lim\{\frac{1}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\}$ – Allonsy Jun 26 '14 at 16:26
• You should edit your question once. Add this into your question. Also specify the limit of $n$, like $\lim_{n\to 0}$ – Saharsh Jun 26 '14 at 16:27
• When it comes to sequences it's always supposed that n goes to infinity :d. – Allonsy Jun 26 '14 at 16:29
• @Allonsy, not always, although maybe so far for you. – Chris K Jun 26 '14 at 16:41

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$2n)/(1+\sqrt n)=0$.

• The limit of the sequence is $\frac12$, since $\frac{n^2}{2n^2+n} \le\{\frac{1}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\} \le \frac{n^2}{2n^2+1}$. The sequences above have limit $\frac12$, and by the squeeze theorem, $\{\frac{1}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\}$ has limit $\frac12$. – Allonsy Jun 26 '14 at 17:05
• Note that I didn't find out that answer, that's what it comes in the book. I don't even know if its true because I can't even see that the original sequence is between that two sequences. – Allonsy Jun 26 '14 at 17:07
• There is no way that is true! – Forever Mozart Jun 26 '14 at 17:08
• I think you wrote the problem incorrectly: look at the numerators. Did you mean for the powers of $n$ to be going from $0$ to $n$ maybe? – Forever Mozart Jun 26 '14 at 17:11
• The first numerator is 1, the second is n, the third is n, the nth is n. The first is 1, the others are n. – Allonsy Jun 26 '14 at 17:12

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: $$\frac1{2+1/n}\le\frac{n}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\le \frac1{2+1/n^2}$$