2
$\begingroup$

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.

  • 0
    Is there anything wrong with my steps above, and also, will doing all of that still lead me to a dead-end?2012-10-24

2 Answers 2

4

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

  • 2
    @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.2012-10-24
3

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.

  • 0
    Nice hint, and identity in your comment above!+12013-02-12