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

  • $\begingroup$ 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

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

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.