Question

Evaluate :-

Answer 1

\[\lim_{n \rightarrow \infty} [\frac{ ( n^2 + 1)( n^2 +1 + 3)(n^2 + 1 + 3 + 5).........2n terms}{ n^(4n) }]^{1/n}\]

Answer 2

@satellite73

Answer 3

Do you mean 2n factors?

Answer 4

It is written 2n terms


@karatechopper

Answer 5

So the last term of the numerator is n^2+1+3+...+(2(2n)-1)?

Answer 6

It's still a factor. Why are we calling it a term? Words mean things.

Answer 7

Sorry bout that.
So the last factor of the numerator is n^2+1+3+...+(2(2n)-1)?

Answer 8

Did anybody else get 1 as the answer?

Answer 9

No Not yet.... I am Weak at Limits.

Answer 10

One logarithm will turn the nasty product into a sum, which is easier to work with.

Answer 11

Plus, the logarithms will allow you to use L'Hospital's Rule easily.

Answer 12

Ho to Solve this Particular Problem.

Answer 13

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: \(n^{2} + n\cdot (n+1) = 2n^2 + n\)

Answer 14

Wait... isn't the sum of the first n odd numbers equal to n^2?

Answer 15

\[\left(\dfrac{n\cdot (n+1)}{2} - n\right)\cdot 2 + n=n(n+1)-2n+n=n^2+n-2n+n=n^2\]

Answer 16

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.

Answer 17

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.

Answer 18

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
\[\lim_{n\to\infty}Q(n)=\lim_{n\to\infty}e^{\ln{Q(n)}}=e^{\lim_{n\to\infty}\ln{Q(n)}}=e^0=1\]

Answer 19

Forgot how to add, I guess.

Answer 20

Is the denominator \(n(4n)\;or\;n^{4n}\)?

Answer 21

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 \(e^{-\infty}=0\).

Answer 22

\[\left[1+\frac{1}{n^2/1}\right]^{n^2}\neq e; \lim_{n\to\infty}\left[1+\frac{1}{n^2/1}\right]^{n^2}=e\]
This means that the limit on the second row should have disappeared after the sums became e's.

Answer 23

Which part of my solution is strange? Taking logarithms is kinda standard for complicated limits of the form \(f(x)^{g(x)}\).

Answer 24

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
\[\lim_{n\to\infty}\left[1+\frac{1}{n^2/1}\right]^{n^2}=e\].

Answer 25

Then is it also true that
\[\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty}1^n\]?

Answer 26

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.

Answer 27

Is this the property you're trying to tell me?
\[\lim_{x\to\infty} f(x)^{g(x)}=\lim_{x\to\infty} \left(\lim_{x\to\infty} f(x)\right)^{g(x)}\]

Answer 28

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?

Answer 29

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.

Answer 30

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.

Answer 31

Naturally, this extends to the sequence with terms \(\frac{1}{n}\ln{P(n)}\) by the statement I made above.

Answer 32

\[\lim_{n\to\infty} \ln{Q(n)}\\
=\lim_{n\to\infty} \frac{1}{n}\left(\ln{(n^2+1)}+\ln{(n^2+4)}+\cdots \ln{(5n^2)}-\ln{n}-\ln{4n}\right)\\
=\lim_{n\to\infty} \frac{1}{n}\ln{(n^2+1)}+\lim_{n\to\infty} \frac{1}{n}\ln{(n^2+4)}+\cdots+ \lim_{n\to\infty}\frac{1}{n}\ln{5n^2}-\lim_{n\to\infty}\frac{1}{n}\ln{n}-\\\lim_{n\to\infty}\frac{1}{n}\ln{4n}\\
=0+0+\cdots+0-0-0=0\]
by my statements above.

Answer 33

I'm going soon but when I return, I'll probably post a numerical analysis of that limit to help us understand it.

Answer 34

Thankyou Sir for the HELP^

Answer 35

Yeah... so I just ran a program to calculate the limit, and it seems like the limit diverges

Answer 36

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;
}