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Need help with some limit
- Thread starter calc_help
- Start date
Write out some of the partial sums (I used Excel):
Code:
n term 1 term 2 term 3 term 4 term 5 term 6 term 7 term 8 term 9 term 10 Sum
1 1.0000 1.000000
2 0.6000 0.5000 1.100000
3 0.4000 0.3846 0.3333 1.117949
4 0.2941 0.3000 0.2800 0.2500 1.124118
5 0.2308 0.2414 0.2353 0.2195 0.2000 1.126955
6 0.1892 0.2000 0.2000 0.1923 0.1803 0.1667 1.128491
7 0.1600 0.1698 0.1724 0.1692 0.1622 0.1529 0.1429 1.129416
8 0.1385 0.1471 0.1507 0.1500 0.1461 0.1400 0.1327 0.1250 1.130016
9 0.1220 0.1294 0.1333 0.1340 0.1321 0.1282 0.1231 0.1172 0.1111 1.130427
10 0.1089 0.1154 0.1193 0.1207 0.1200 0.1176 0.1141 0.1098 0.1050 0.1000 1.130721What Dr. Phil says is true--this is what you are doing wrong. But what should you be doing to get it right? Here is a hint: split the problem into two pieces,
sum (n/(n^2+i^2) + sum(i/(n^2+i^2)
Now compare each of these sums to a corresponding integral from 1 to n or 0 to n-1 according to whether you need an over estimate or underestimate, for instance
integral from 1 to n (n/(n^2 + x^2) ) dx
for the first sum. In each case the integrals can be evaluated, and in each case since the integrand is either decreasing (first sum) or increasing (second sum), you can get comparisons between the integral and the sum.
sum (n/(n^2+i^2) + sum(i/(n^2+i^2)
Now compare each of these sums to a corresponding integral from 1 to n or 0 to n-1 according to whether you need an over estimate or underestimate, for instance
integral from 1 to n (n/(n^2 + x^2) ) dx
for the first sum. In each case the integrals can be evaluated, and in each case since the integrand is either decreasing (first sum) or increasing (second sum), you can get comparisons between the integral and the sum.
Let \(\displaystyle f(x) = \dfrac{1}{1+x^2}\) and \(\displaystyle g(x) = \dfrac{x}{1+x^2}\)
then note that
\(\displaystyle \displaystyle \int_0^1 f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(\frac{i}{n})\cdot\frac{1}{n}\)
\(\displaystyle \displaystyle\int_0^1 g(x) dx = \lim_{n\to\infty}\sum_{i=1}^n g(\frac{i}{n})\cdot\frac{1}{n}\)
S your limit is \(\displaystyle \int_0^1 (f(x)+g(x))dx\)
then note that
\(\displaystyle \displaystyle \int_0^1 f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(\frac{i}{n})\cdot\frac{1}{n}\)
\(\displaystyle \displaystyle\int_0^1 g(x) dx = \lim_{n\to\infty}\sum_{i=1}^n g(\frac{i}{n})\cdot\frac{1}{n}\)
S your limit is \(\displaystyle \int_0^1 (f(x)+g(x))dx\)
I'm studying this stuff myself, so could be wrong.
So basically, after plugging in increasing x values (in increasing "n = 1, 2, 3 etc.." sequence form) and reducing, you got 1/1, 1/2, 1/3, etc... with the denominator gradually getting larger. In this math study book I bought, it gives a similar example and the limit is 0. That's because an increasing denominator divided by 1 gets gradually smaller (approaches 0). Of course, since this is an increasing "n" sequence fraction, what happens in the denominator is proportional to what happens when the fraction is divided.
But then again, this is a "partial sum problem" so that doesn't apply here. Perhaps, the limit is some repeating number (that comes after adding up all the divided fractions).
So basically, after plugging in increasing x values (in increasing "n = 1, 2, 3 etc.." sequence form) and reducing, you got 1/1, 1/2, 1/3, etc... with the denominator gradually getting larger. In this math study book I bought, it gives a similar example and the limit is 0. That's because an increasing denominator divided by 1 gets gradually smaller (approaches 0). Of course, since this is an increasing "n" sequence fraction, what happens in the denominator is proportional to what happens when the fraction is divided.
But then again, this is a "partial sum problem" so that doesn't apply here. Perhaps, the limit is some repeating number (that comes after adding up all the divided fractions).
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The individual terms go to 0 in the limit, but the sum does not. Sums of such series are frequently irrational numbers involving pi and e. In this case, each of the two integrals shown by daon2 gives an irrational number. I was amused to see how rapidly the numerical sequence I gave is approaching the limit, which is 1.13197...I'm studying this stuff myself, so could be wrong.
So basically, after plugging in increasing x values (in increasing "n = 1, 2, 3 etc.." sequence form) and reducing, you got 1/1, 1/2, 1/3, etc... with the denominator gradually getting larger. In this math study book I bought, it gives a similar example and the limit is 0. That's because an increasing denominator divided by 1 gets gradually smaller (approaches 0). Of course, since this is an increasing "n" sequence fraction, what happens in the denominator is proportional to what happens when the fraction is divided.
But then again, this is a "partial sum problem" so that doesn't apply here. Perhaps, the limit is some repeating number (that comes after adding up all the divided fractions).
1 ) That's because an increasing denominator divided by 1
2) gets gradually smaller (approaches 0).
1) This is backwards. Use
"an increasing denominator divided into 1," or
"1 divided by an increasing denominator," for example.
2) I'm going to put the issue of using the subjective
word "gradually" aside.
The parenthetical phrase, "approaches 0," is not equivalent
to your stated idea of fractions with increasing denominators
divided [into] 1 getting gradually smaller.
Here is a counterexample of a sequence that has "gradually" increasing
denominators, but successive terms are approaching 1/2, not 0:
\(\displaystyle \dfrac{1}{1.9}, \ \dfrac{1}{1.99}, \ \dfrac{1}{1.999}, \ \dfrac{1}{1.9999}, \ ...\)