I think it is related to squeeze theorem, but could not come up with a solution. The answer here is $1-\ln(9/4)$. Can someone help me with this question? $$ \lim_{n\to\infty}\sum_{k=1}^{n} \frac{k}{k\,n+2n^2} = \lim_{n \to \infty} \left(\frac{1}{n+2n^2} +\frac{2}{2n+2n^2}+\cdots + \frac{n}{n^2+2n^2}\right) $$ Any help will be appreciated! Thanks!

  • $\begingroup$ Intuitively, it should go to 0,but that's not a proof.According to you,it DOESN'T go to 0. $\endgroup$ – Mathemagician1234 Feb 20 '15 at 6:29

Hint: This is the limit of a Riemann sum.

$$\lim_{n\to\infty}\sum_{k=1}^{n} \frac{k}{k\,n+2n^2} = \lim_{n\to\infty}\frac1{n}\sum_{k=1}^{n} \frac{k/n}{k/n+2}= \,\,\ldots$$

  • $\begingroup$ Ok,you baffled me-how did you guess that? Riemann sum was the last thing I was thinking. $\endgroup$ – Mathemagician1234 Feb 20 '15 at 6:46
  • $\begingroup$ Ok,now you HAVE to tell me how you got that. Email it to me if you don't want to blow it for the OP. $\endgroup$ – Mathemagician1234 Feb 20 '15 at 6:48
  • $\begingroup$ Riemann sum was the first thing I was thinking, but only because I've recently taken the Mathematics GRE... $\endgroup$ – Gyu Eun Lee Feb 20 '15 at 6:48
  • $\begingroup$ I have to seriously review computng sums and series,I am WAY out of practice.The old British analysis texts,like Ferrar, are really good for that. $\endgroup$ – Mathemagician1234 Feb 20 '15 at 6:49
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    $\begingroup$ Any time I see a limit of a partial sum with $1/n$ in front I automatically l0ok for a Riemann sum $\endgroup$ – RRL Feb 20 '15 at 6:51

Thank RRL for contributing the Riemann Sum solution... This is what I get for this question as a result.

You divide function $\frac{x}{x+2}$ into n pieces on interval [0,1], so that each small interval (A little confused about why [0,1] is chosen to make it work, rather than [1,2]) $$\Delta x=\frac1{n}$$ so the Riemann sum of this function becomes, \begin{align*} &\qquad\frac1{n}\left(f(\frac1{n})+f(\frac{2}{n})+f(\frac3{n})+\cdots+f(\frac{n}{n})\right)\\ &=\frac1{n}\left(\frac{1/n}{1/n+2}+\frac{2/n}{2/n+2}+\cdots+\frac{n/n}{n/n+2}\right)\\ &=\frac1{n}\left(\sum_{k=1}^{n}\frac{k/n}{k/n+2}\right) \end{align*} which means we have to calculate the following integral ultimately, $$\int _0 ^ 1 \frac{x}{x+2}dx$$

Thank other people for contributing their ideas as well!

  • $\begingroup$ I do not know if this is a correct way to get the final answer, though. $\endgroup$ – Terrence Liao Feb 20 '15 at 8:30
  • $\begingroup$ It's correct. $\int_0^1 \frac{x}{x+2}dx = \int_0^1 \left[1-\frac{2}{x+2}\right]dx = [x - 2 \ln(x+2)]_0^1 = 1 - \ln(9/4)$ $\endgroup$ – RRL Feb 20 '15 at 8:40
  • $\begingroup$ why choosing [0,1] rather than [1,2] could make it work? [1,2] also gives me subinterval of $\Delta x = (2-1)/n$, doesn't it? $\endgroup$ – Terrence Liao Feb 20 '15 at 19:30
  • $\begingroup$ The length of each subinterval is $1/n$, but the partition points $1/n,2/n,\ldots,1$ where the function is evaluated are in the interval $[0,1]$ -- so this sum converges to the integral over $[0,1]$. $\endgroup$ – RRL Feb 20 '15 at 19:59

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