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I want to find

$$\lim_{n \rightarrow \infty} \left(\frac{n + 1}{n^2 + 1} + \frac{n + 2}{n^2 + 2} + \cdots + \frac{n + n}{n^2 + n} \right)$$

Here is my solution.

$$\lim_{n \rightarrow \infty} \left(\frac{n + 1}{n^2 + 1} + \frac{n + 2}{n^2 + 2} + \cdots + \frac{n + n}{n^2 + n} \right) = \lim_{n \rightarrow \infty} \left(\frac{n}{n^2 + 1} + \frac{n}{n^2 + 2} + \cdots + \frac{n}{n^2 + n} \right) + \left(\frac{1}{n^2 + 1} + \frac{2}{n^2 + 2} + \cdots + \frac{n}{n^2 + n} \right)$$

Lets call the first sum $S^1_n$ and the second sum $S^2_n$. Then we have:

$$1 \leftarrow n \frac{n}{n^2 + n} \leq S^1_n \leq n \frac{n}{n^2 + 1} \rightarrow 1$$

And by the sandwich criterium $S^1_n \rightarrow 1$.

On the other hand we have:

$$ \frac{1}{2} \leftarrow \frac{\frac{n (n + 1)}{2}}{n^2 + n} = \frac{1 + \cdots + n}{n^2 + n} \leq S^2_n \leq \frac{1 + \cdots + n}{n^2 + 1} = \frac{\frac{n (n + 1)}{2}}{n^2 + 1} \rightarrow \frac{1}{2}$$

So by the sandwich criterium $S^2_n \rightarrow \frac{1}{2}$.

So the limit is $\frac{3}{2}$.

Is this correct? Are there any other alternatives?

Jordi
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    It looks to be correct; as a purely stylistic suggestion you could have done both of the manipulations simultaneously, bounding $S_n$ between $\left(n^2 + \frac{n(n+1)}{2}\right)/(n^2+n)$ and $\left(n^2 + \frac{n(n+1)}{2}\right)/(n^2+1)$. – HehBot Feb 02 '24 at 10:27
  • I have closed the question as a duplicate of https://math.stackexchange.com/q/1075305/42969 because that question is about the same limit and has three different solutions, including one that is essentially your approach. – Martin R Feb 02 '24 at 11:01

1 Answers1

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As noticed in the comments, your proof looks fine.

As an alternative, we can use Riemann sum, noting that

$$\sum_{i=1}^n\frac{n + i}{n^2 + i}=\frac1n\sum_{i=1}^n\frac{1 + \frac i n}{1 + \frac i{n^2}} \to \int_0^1(1+x)dx =\left[x+\frac{x^2}2\right]_0^1=\frac32$$

where we have neglected, in a heuristic approch, the term $i/n^2$ and which needs to be made more rigorous as indicated in the related/duplicate question

user
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    Not obvious that the limit is the same to $\frac{1}{n}\sum_{i=1}^n 1+\frac{i}{n}$, that is Riemann sum. It is top estimation, of course, but no reasons for being equal – Egor Ivanov Feb 02 '24 at 10:48
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    Can you explain, please, better your answer? I have not found the proof of the $\sum_{i=1}^n\frac{n + i}{n^2 + i}=\frac1n\sum_{i=1}^n\frac{1 + \frac i n}{1 + \frac i{n^2}} $ and why do you consider the integral between from 0 to 1? – Sebastiano Feb 02 '24 at 10:50
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    @EgorIvanov Yes your right, it requires some more effort and discussion to make it rigorous. I add a commento on that. Thanks – user Feb 02 '24 at 10:54
  • Riemann sum solution already here: https://math.stackexchange.com/a/1075321/42969 – Martin R Feb 02 '24 at 10:54
  • @EgorIvanov I'm not sure whether or not in this form the answer can be useful or it might be prefereble delete it. Thanks – user Feb 02 '24 at 11:03
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    @Sebastiano This is an application of riemann sum with the "lack" already noticed by Egor which requires to justify the assumption to neglect the $i/n^2$ term. – user Feb 02 '24 at 11:07
  • @user It is an interesting approach to obtain the top estimation, I think you shouldn't delete it – Egor Ivanov Feb 02 '24 at 11:18
  • @MartinR For an another alternative, you suggest to add it here or in the duplicate? – user Feb 02 '24 at 11:42
  • @user: I suggest to add all new/alternative answers to the older thread, and to delete it here. – Martin R Feb 02 '24 at 11:56
  • @user Ah ok. I have understood. thanks. – Sebastiano Feb 02 '24 at 19:20