Evaluate :-

- goformit100

Evaluate :-

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- 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

@satellite73

- tkhunny

Do you mean 2n factors?

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## More answers

- goformit100

It is written 2n terms
@karatechopper

- blockcolder

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

- tkhunny

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

- blockcolder

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

- blockcolder

Did anybody else get 1 as the answer?

- goformit100

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

- blockcolder

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

- blockcolder

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

- goformit100

Ho to Solve this Particular Problem.

- 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

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

- 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

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

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

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

Forgot how to add, I guess.

- tkhunny

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

- 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

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

- blockcolder

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

- 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

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

- 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

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

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

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

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

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

- 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

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

- goformit100

Thankyou Sir for the HELP^

- anonymous

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

- 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;
}

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