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#### QuestForInsight

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- Jul 22, 2012

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Show that $$ \lim_{n \to \infty} \prod_{1 \le k \le n}\left(1+\frac{kx}{n^2}\right) = e^{x/2}$$

- Thread starter QuestForInsight
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- #1

- Jul 22, 2012

- 35

Show that $$ \lim_{n \to \infty} \prod_{1 \le k \le n}\left(1+\frac{kx}{n^2}\right) = e^{x/2}$$

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- #2

$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{kx}{n^2} \right)=L$

$\displaystyle \lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+ \frac{kx}{n^2} \right) \right)=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left(1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{kx}{n^2}=\ln(L)$

$\displaystyle x\lim_{n\to\infty}\frac{n^2+n}{2n^2}=\ln(L)$

$\displaystyle \frac{x}{2}=\ln(L)$

$\displaystyle L=e^{\frac{x}{2}}$

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- #3

- Jul 22, 2012

- 35

Brilliant!

It's much easier than the solution I know.

It's much easier than the solution I know.

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- #4

- Feb 5, 2012

- 1,621

Hi MarkFL,

$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{kx}{n^2} \right)=L$

$\displaystyle \lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+ \frac{kx}{n^2} \right) \right)=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left(1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{kx}{n^2}=\ln(L)$

$\displaystyle x\lim_{n\to\infty}\frac{n^2+n}{2n^2}=\ln(L)$

$\displaystyle \frac{x}{2}=\ln(L)$

$\displaystyle L=e^{\frac{x}{2}}$

I 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.

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- #6

- Feb 7, 2012

- 2,705

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 ).My method tacitly assumes the limit exists. I suspect your method is more rigorous.

Start by using the Taylor expansion of $\ln(1+t)$ to deduce that $\ln(1+t) = t + O(t^2)$ as $t\to0$. Then $$\begin{aligned}\sum_{k=1}^n \ln\Bigl(1 + \frac{kx}{n^2}\Bigr) &= \sum_{k=1}^n \Bigl(\frac{kx}{n^2} + O\bigl(\tfrac{k^2}{n^4}\bigr)\Bigr) \\ &= \frac{\frac12n(n+1)x}{n^2} + n\cdot O\bigl(\tfrac1{n^2}\bigr) \\ &= \tfrac12x + O\bigl(\tfrac1{n}\bigr). \end{aligned} $$ Now let $n\to\infty$ to get $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n \ln\Bigl(1 + \frac{kx}{n^2}\Bigr) = \frac x2$ and hence (because the exponential function is continuous) $\displaystyle \lim_{n \to \infty} \prod_{k=1}^n \Bigl(1+\frac{kx}{n^2}\Bigr) = e^{x/2}.$

I'm not entirely confident about that argument because it's all too easy to use the "big O" notation to disguise lapses in rigour. But it looks to me as though it's basically correct.

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- #7

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.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.

- Feb 5, 2012

- 1,621

That is true but note the the summation includes \(n\) as the upper limit. In that case I don't think that this property generally holds. For example,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.

\[2=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}2^{k-n}\neq\sum_{k=1}^{n}\lim_{n\rightarrow\infty}2^{k-n}=0\]

Kind Regards,

Sudharaka.

- Admin
- #9

Rather than edit my initial post, I will rewrite it so that others at my level might see the mistake and learn from it as well.

We are given:

$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{kx}{n^2} \right)=L$

$\displaystyle \lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+ \frac{kx}{n^2} \right) \right)=\ln(L)$

$\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{\ln\left(\left(1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)$

$\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{kx}{n^2}=\ln(L)$

$\displaystyle x\lim_{n\to\infty}\frac{n^2+n}{2n^2}=\ln(L)$

$\displaystyle \frac{x}{2}=\ln(L)$

$\displaystyle L=e^{\frac{x}{2}}$

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- #10

- Jul 22, 2012

- 35

In your rewrite, I'm not sure how you reduced the summand to $\frac{kx}{n^2}$ without interchanging the sum and the limit, though?

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- #11

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- #12

- Jul 22, 2012

- 35

$ \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 $\displaystyle \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) = e^{x/2}$.

- Feb 5, 2012

- 1,621

Hi QuestForInsight,

$ \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 $\displaystyle \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) = e^{x/2}$.

Thank you for the solution. I think this proof is valid only for \(x>-1\) since the Bernoulli's inequality is defined as such. Isn't?

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- #14

- Feb 29, 2012

- 342

You cannot assume the existence of certain objects

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- #16

As I say, if the implications reverse then it is both neccesary and sufficent that the answer is the limit. In plain language, it is the limit.

You cannot assume the existence of certain objectsa priori, that's why the argument was labeled as sloppy.

- Feb 15, 2012

- 1,967

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

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- #19

- Feb 7, 2012

- 2,705

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 $x$.

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

- Feb 15, 2012

- 1,967

and it is important for post #6, because of the comparatively small region of convergence for the taylor series of log(1+t) (presumably centered at 0).

the convergence is important (in my mind, anyway) so that we know the big O terms can indeed be disregarded.

and yes, that does avoid the limit/sum interchange the proof of which seems to me problemmatic.

so far, your (Opalg's) proof is the only one that "makes sense" to me.

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.

- Feb 5, 2012

- 1,621

Hi Deveno,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.

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.

- Feb 15, 2012

- 1,967

thanks for the link, yes, since k (the index) is bounded by by n, 0 ≤ k/n ≤ 1, which is what i was missing.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.