Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Is there anything wrong with my steps above, and also, will doing all of that still lead me to a dead-end? –  Yellow Skies Oct 24 '12 at 18:14
add comment

2 Answers

up vote 2 down vote accepted

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

share|improve this answer
    
Wow! How did you know when to use it? Instinct? Just wondering... –  Yellow Skies Oct 24 '12 at 18:09
    
@SingaporeanDude.: He feels it.This is beyond the instinct. ;-) –  B. S. Oct 24 '12 at 18:12
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. –  André Nicolas Oct 24 '12 at 18:12
add comment

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.

share|improve this answer
    
How did you pull out the values for the upper and lower bound? as well as f(x) –  Yellow Skies Oct 24 '12 at 18:11
1  
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. :) –  B. S. Oct 24 '12 at 18:18
    
Thanks!!!!!!!!!!! –  Yellow Skies Oct 24 '12 at 18:22
    
Nice hint, and identity in your comment above!+1 –  amWhy Feb 12 '13 at 0:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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