Find the limit $$\lim_{n\to \infty}\sum_{i=1}^{n}\frac{i}{n^2+i^2}$$ by expressing it as a definite integral of an appropiate function via Riemann Sums.

Observation: $n$ must refer to the number of slices, and $i$ must refer to $i$th slice.

My attempt.

First I revisited Riemann Sums. Assume what I am trying to find have the form $\int_{a}^{b}f(x).$ Cutting up the bound $(a,b)$ into $n$ slices, the length of each piece with respect to $x$ is $\frac{b-a}{n}. $

Next, looking at each slice, I decide to take the right hand value for convenience, that is $a+i\frac{b-a}{n}. $ Now, I clearly have the area of each slice, that is, $$\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$$

And when I sum up all of the slices, I have $$\sum_{i=1}^{n}\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$$

And finally, increasing the number of cuts to make the area as accurate as possible, we have $$\lim_{n\to \infty}\sum_{i=1}^{n}\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$$

Therefore I conclude that $$\frac{i}{n^2+i^2}=\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$$

Whats left now is to find b,a and f(x). After all the work, I feel closer to my answer, yet so far away from it.

Any hints? Thanks in advance! List them as solutions. I am looking for hints only.


marked as duplicate by YuiTo Cheng, Lee David Chung Lin, Shogun, Lord Shark the Unknown, Cesareo Jun 29 at 7:41

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  • $\begingroup$ Is there anything wrong with my steps above, and also, will doing all of that still lead me to a dead-end? $\endgroup$ – Yellow Skies Oct 24 '12 at 18:14
  • $\begingroup$ @YuiToCheng Why make this older post the duplicate? The content in the newer post isn't necessarily better. $\endgroup$ – Lee David Chung Lin Jun 29 at 0:56

Hint: Divide top and bottom by $n^2$, expressing the result as $$\frac{1}{n}\sum_1^n \frac{i/n}{1+i^2/n^2}.$$

  • $\begingroup$ Wow! How did you know when to use it? Instinct? Just wondering... $\endgroup$ – Yellow Skies Oct 24 '12 at 18:09
  • 1
    $\begingroup$ @SingaporeanDude.: He feels it.This is beyond the instinct. ;-) $\endgroup$ – mrs Oct 24 '12 at 18:12
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    $\begingroup$ @SingaporeanDude.: It is no mystery, I have seen something of the same general character before! But it is not too hard to get to it. We want a $\dfrac{1}{n}$ in front. $\endgroup$ – André Nicolas Oct 24 '12 at 18:12

Hint: take $a=0, b=1$ and $f(x)=\frac{1}{x+\frac{1}{x}}=\frac{x}{1+x^2}$. This is consistent to Andre's answer.

  • $\begingroup$ How did you pull out the values for the upper and lower bound? as well as f(x) $\endgroup$ – Yellow Skies Oct 24 '12 at 18:11
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    $\begingroup$ Just consider the identity you pointed $$\frac{i}{n^2+i^2}=\frac{b-a}{n}\times f\bigg(a+i\frac{b-a}{n}\bigg )$$ and think of the possible values for $a$ and $b$. Of course some manipulations like what Andre noted needed here. Try to do for another summand by yourself. You can do it. That is it. :) $\endgroup$ – mrs Oct 24 '12 at 18:18
  • $\begingroup$ Nice hint, and identity in your comment above!+1 $\endgroup$ – Namaste Feb 12 '13 at 0:07

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