- #1
QuestForInsight
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Show that $$ \lim_{n \to \infty} \prod_{1 \le k \le n}\left(1+\frac{kx}{n^2}\right) = e^{x/2}$$
MarkFL said:We are given:
$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{kx}{n^2} \right)=L$
$\displaystyle \lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+ \frac{kx}{n^2} \right) \right)=\ln(L)$
$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left(1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)$
$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{kx}{n^2}=\ln(L)$
$\displaystyle x\lim_{n\to\infty}\frac{n^2+n}{2n^2}=\ln(L)$
$\displaystyle \frac{x}{2}=\ln(L)$
$\displaystyle L=e^{\frac{x}{2}}$
I suspect that your argument can be modified to avoid the assumption that the limit exists (and also avoiding the fancy footwork with interchanging limits and taking partial limits (Wondering)).MarkFL said:My method tacitly assumes the limit exists. I suspect your method is more rigorous.
Sudharaka said:Hi MarkFL, :)
I don't know if I am missing something here, but I don't quite understand how you got,
\[\sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left( 1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)\]
from,
\[\lim_{n\to\infty}\sum_{k=1}^n\frac{\ln\left(\left( 1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)\]
Can you please elaborate? :)
Kind Regards,
Sudharaka.
MarkFL said:I suppose it was unnecessary to take the summation outside the limit, but I was using the property that the limit of a sum is the sum of the limits.
QuestForInsight said:Using Bernoulli's Inequality we have:
$ \begin{aligned} \displaystyle & \hspace{0.5in}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \bigg(1+\frac{kx}{n^2}\bigg) \le \bigg(1+\frac{x}{n^2}\bigg)^k \\& \iff \prod_{ 0\le k \le n}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \prod_{ 0\le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \prod_{ 0\le k \le n}\bigg(1+\frac{x}{n^2}\bigg)^k \\& \iff \hspace{0.1in} \bigg(1+\frac{x}{n}\bigg)^{ \frac{1}{2}(n+1)} \le \prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \bigg(1+\frac{x}{n^2}\bigg)^{ \frac{1}{2} n(n+1)} \\& \iff \lim_{n \to \infty}\bigg(1+\frac{x}{n}\bigg)^{ \frac{1}{2}(n+1)} \le \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \lim_{n \to \infty}\bigg(1+\frac{x}{n^2}\bigg)^{ \frac{1}{2} n(n+1)} \\& \iff e^{x/2} \le \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le e^{x/2}.\end{aligned}$
Therefore $\displaystyle \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) = e^{x/2}$.
Fantini said:What if the limit didn't exist at all, and you arrived at such 'solution'? It may seem a trivial matter, but beforehand you couldn't affirm that the limit exists. What if the question was posed as "Find out whether the limit exists or not"?
You cannot assume the existence of certain objects a priori, that's why the argument was labeled as sloppy. (Wondering)
When looking at the limit (if it exists) $\displaystyle \lim_{n \to \infty} \prod_{k=1}^n \Bigl(1+\frac{kx}{n^2}\Bigr)$, we are only interested in what happens when $n$ gets large. So we might as well assume from the start that we are only going to look at values of $n$ large enough to ensure that $|x/n|$ is small. In particular, if $|x/n|<1$ then $1-\dfrac{kx}{n^2}>0$ (for $1\leqslant k\leqslant n$), and so it will be legitimate to take its logarithm. If you then use the argument in my post #6, you can check that the limit exists and equals $e^{\,x/2}$, for all real $x$.Deveno said:please excuse my denseness, but i am troubled by a few points.
first of all, for which x is this supposed to hold? the reason i ask is because of the questions raised by Sudharaka regarding the Bernoulli's inequality proof, and because of MarkFL's liberal use of the logarithm function (which is only defined for positive real numbers).
i am also troubled by the point raised in post #8. i can't see that this objection has been fully answered. to justify a method by "it gives the right answer" seems poor reasoning:
after all, 2*2 = 2+2, but we cannot conclude from this that if we need to square a number, we can get by with doubling it.
i am also troubled by the inequalities in line 2 of post #12, they seem to tacitly assume each term in the n-fold products is non-negative (am i missing something?). i fear things may go very badly if x = -100, for example.
i understand the general strategy here: we want to leverage a known limit for e into something we can use. since both sides of the originally posted limit make sense for all real x, i would feel better about a proof which does not restrict what x may be (in all fairness, some of these concerns might be allayed by requiring n be at least large enough to ensure my concerns are invalid, but this point seems glossed over in the preceding posts).
please enlighten me, for my own peace of mind :)
Deveno said:i am also puzzled by one of the inequalities in QuestForInsight's post (#12). try as i might, i cannot obtain the lower bound, and it seems to me that bernoulli's inequality actually tells us:
\(\displaystyle \left(1 + \frac{kx}{n^2}\right) \leq \left(1 + \frac{x}{n}\right)^{\frac{k}{n}}\)
again, i may just be old and senile, but i find this particular problem rather interesting, and would like to understand it better.
Sudharaka said:Hi Deveno, :)
I think that the inequality in post #12 is correct. We have to use the Generalized Bernoulli's inequality to show that.
Kind Regards,
Sudharaka.
The exponential limit is the value that a function approaches as its input approaches a certain value, usually infinity or negative infinity.
The formula for calculating exponential limit is: lim x->a f(x) = L, where L is the exponential limit of the function f(x) as x approaches the value a.
To prove the exponential limit of a function, you must show that as the input of the function approaches a certain value, the output of the function approaches a specific value, which is the exponential limit. This can be done through various mathematical techniques such as substitution, algebraic manipulation, and using the definition of limit.
The exponential limit of e^(x/2) is 1. This can be shown by substituting different values for x as it approaches infinity, and observing that the output approaches 1.
Yes, the exponential limit of e^(x/2) is always 1, regardless of the value of x. This is because the function e^(x/2) approaches 1 as x approaches infinity or negative infinity, and the exponential limit is defined as the value that the function approaches at these points.