goformit100
  • goformit100
Evaluate :-
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
goformit100
  • goformit100
\[\lim_{n \rightarrow \infty} [\frac{ ( n^2 + 1)( n^2 +1 + 3)(n^2 + 1 + 3 + 5).........2n terms}{ n^(4n) }]^{1/n}\]
goformit100
  • goformit100
@satellite73
tkhunny
  • tkhunny
Do you mean 2n factors?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

goformit100
  • goformit100
It is written 2n terms @karatechopper
blockcolder
  • blockcolder
So the last term of the numerator is n^2+1+3+...+(2(2n)-1)?
tkhunny
  • tkhunny
It's still a factor. Why are we calling it a term? Words mean things.
blockcolder
  • blockcolder
Sorry bout that. So the last factor of the numerator is n^2+1+3+...+(2(2n)-1)?
blockcolder
  • blockcolder
Did anybody else get 1 as the answer?
goformit100
  • goformit100
No Not yet.... I am Weak at Limits.
blockcolder
  • blockcolder
One logarithm will turn the nasty product into a sum, which is easier to work with.
blockcolder
  • blockcolder
Plus, the logarithms will allow you to use L'Hospital's Rule easily.
goformit100
  • goformit100
Ho to Solve this Particular Problem.
tkhunny
  • tkhunny
First off, write the factors rationally. That's just crazy 1 + 3 + 5 + ...+ (2n-1) - This is the sum of the first n odd natural numbers. Break it up a little: 0 + 2 + 4 + ...+ (2n-2) + n - Just stealing one from each term and putting them all back on the end. (0 + 1 + 2 + ...+ (n-1))*2 + n - A little factoring. What's the sum of the first n-1 Natural Numbers? \(\dfrac{n\cdot (n+1)}{2} - n\) Finally, we have an expression for the whole silly set of odd Natural Numbers. \(\left(\dfrac{n\cdot (n+1)}{2} - n\right)\cdot 2 + n\) This simplifies nicely. and we see that each factor is now: \(n^{2} + n\cdot (n+1) = 2n^2 + n\)
blockcolder
  • blockcolder
Wait... isn't the sum of the first n odd numbers equal to n^2?
blockcolder
  • blockcolder
\[\left(\dfrac{n\cdot (n+1)}{2} - n\right)\cdot 2 + n=n(n+1)-2n+n=n^2+n-2n+n=n^2\]
blockcolder
  • blockcolder
Anyway, the point is that the numerator can be simplified into \((n^2+1)(n^2+4)\cdots (n^2+n^2)\). Take the logarithm of the expression whose limit we are taking: \[\ln{\left[\frac{(n^2+1)(n^2+4)\cdots (n^2+n^2)}{n(4n)}\right]^{1/n}\\=\frac{1}{n}}\left(\ln{(n^2+1)}+\ln{(n^2+4)}+\cdots \ln{(2n^2)}-\ln{n}-\ln{4n}\right)\] Let's examine each term carefully.
blockcolder
  • blockcolder
After distributing, each term is of the form \(\frac{1}{n}\ln{P(n)}=\frac{\ln{P(n)}}{n}\), where \(P(n)=a_kn^k+a_{k-1}x^{k-1}+\cdots+a_1n+a_0.\) \[\lim_{n\to\infty}\frac{1}{n}\ln{P(n)}=\lim_{n\to\infty}\frac{\ln{P(n)}}{n}=\lim_{n\to\infty}\frac{P'(n)}{P(n)}\] after applying L'Hospital's Rule.
blockcolder
  • blockcolder
This last limit is clearly equal to 0, because the degree of P'(x) is less than the degree of P(n). Thus, \[\lim_{n\to\infty}\ln{\left[\frac{(n^2+1)(n^2+4)\cdots (n^2+n^2)}{n(4n)}\right]^{1/n}}=0\] From here, if you let \[Q(n)=\left[\frac{(n^2+1)(n^2+4)\cdots (n^2+n^2)}{n(4n)}\right]^{1/n}\] then \[\lim_{n\to\infty}Q(n)=\lim_{n\to\infty}e^{\ln{Q(n)}}=e^{\lim_{n\to\infty}\ln{Q(n)}}=e^0=1\]
tkhunny
  • tkhunny
Forgot how to add, I guess.
tkhunny
  • tkhunny
Is the denominator \(n(4n)\;or\;n^{4n}\)?
blockcolder
  • blockcolder
If it's n(4n), the answer is 1; if it's n^(4n), it's 0. What would change in my solution if it's n^(4n) is that the last term of the sum above is \(-4n\ln{n}\) instead of \(-\ln{n}-\ln{4n}\). Thus, after expanding, the last term is \(\frac{1}{n}\cdot -4n\ln{n}=-4\ln{n}\) and thus, \(\lim_{n\to\infty} \ln{Q(n)}=-\infty\) and \(e^{-\infty}=0\).
blockcolder
  • blockcolder
\[\left[1+\frac{1}{n^2/1}\right]^{n^2}\neq e; \lim_{n\to\infty}\left[1+\frac{1}{n^2/1}\right]^{n^2}=e\] This means that the limit on the second row should have disappeared after the sums became e's.
blockcolder
  • blockcolder
Which part of my solution is strange? Taking logarithms is kinda standard for complicated limits of the form \(f(x)^{g(x)}\).
blockcolder
  • blockcolder
My point is that you can't really substitute e for \[\left[1+\frac{1}{n^2/1}\right]^{n^2}\] because it is not true that \[\left[1+\frac{1}{n^2/1}\right]^{n^2}=e.\] What IS true is that \[\lim_{n\to\infty}\left[1+\frac{1}{n^2/1}\right]^{n^2}=e\].
blockcolder
  • blockcolder
Then is it also true that \[\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty}1^n\]?
blockcolder
  • blockcolder
Then please tell me where I went wrong in my solution because I believe in my solution. If you can show me where I made an error, I'll concede that your answer is the right one.
blockcolder
  • blockcolder
Is this the property you're trying to tell me? \[\lim_{x\to\infty} f(x)^{g(x)}=\lim_{x\to\infty} \left(\lim_{x\to\infty} f(x)\right)^{g(x)}\]
blockcolder
  • blockcolder
Yeah, L'Hospital's rule can't be used on sequences. However, it is true that if \(\lim_{x\to\infty}f(x)=L\) and \(f(n)=a_n\) for integers n, then \(\lim_{n\to\infty}a_n=L\). Thus, I applied L'Hospital's rule on the function f(x), and that is allowed, isn't it?
blockcolder
  • blockcolder
So it's technically my fault that I forgot that detail, but still, if you replace all instances of n with x in my solution, then you can apply the last statement that I said and reach my conclusion.
blockcolder
  • blockcolder
That is irrelevant since I had proved that \[\lim_{x\to\infty}\frac{1}{x}\ln{(P(x)})=0\] for all polynomials P(x), so it doesn't matter whether it's n^2+4n^2 or n^2+500n^2 (an exaggeration, but I hope you get what I mean) since they're all polynomials.
blockcolder
  • blockcolder
Naturally, this extends to the sequence with terms \(\frac{1}{n}\ln{P(n)}\) by the statement I made above.
blockcolder
  • blockcolder
\[\lim_{n\to\infty} \ln{Q(n)}\\ =\lim_{n\to\infty} \frac{1}{n}\left(\ln{(n^2+1)}+\ln{(n^2+4)}+\cdots \ln{(5n^2)}-\ln{n}-\ln{4n}\right)\\ =\lim_{n\to\infty} \frac{1}{n}\ln{(n^2+1)}+\lim_{n\to\infty} \frac{1}{n}\ln{(n^2+4)}+\cdots+ \lim_{n\to\infty}\frac{1}{n}\ln{5n^2}-\lim_{n\to\infty}\frac{1}{n}\ln{n}-\\\lim_{n\to\infty}\frac{1}{n}\ln{4n}\\ =0+0+\cdots+0-0-0=0\] by my statements above.
blockcolder
  • blockcolder
I'm going soon but when I return, I'll probably post a numerical analysis of that limit to help us understand it.
goformit100
  • goformit100
Thankyou Sir for the HELP^
anonymous
  • anonymous
Yeah... so I just ran a program to calculate the limit, and it seems like the limit diverges
anonymous
  • anonymous
static double n; public static void main(String[] args) { Scanner oScan = new Scanner(System.in); n = 1; while(true){ System.out.println(calculate()); n *= 2; oScan.nextLine(); } } public static double calculate(){ double result = 1; result /= 4*n*n; for(int i = 0; i<2*n; i++){ result *= getTerm(i+1); } result = Math.pow(result, 1/n); return result; } public static double getTerm(double i){ return n*n+i*i; }

Looking for something else?

Not the answer you are looking for? Search for more explanations.