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

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

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  • $\begingroup$ 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 $\frac12$. $\endgroup$ – Allonsy Jun 26 '14 at 17:05
  • $\begingroup$ 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. $\endgroup$ – Allonsy Jun 26 '14 at 17:07
  • $\begingroup$ There is no way that is true! $\endgroup$ – Forever Mozart Jun 26 '14 at 17:08
  • $\begingroup$ 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? $\endgroup$ – Forever Mozart Jun 26 '14 at 17:11
  • $\begingroup$ 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. $\endgroup$ – Allonsy Jun 26 '14 at 17:12
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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: $$ \frac1{2+1/n}\le\frac{n}{2n^2+1}+\frac{n}{2n^2+2}+\dots+\frac{n}{2n^2+n}\le \frac1{2+1/n^2} $$

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