3
$\begingroup$

I realize that this question already has an answer elsewhere and now realize what the correct answer is (substituting into the limit form of the exponential function and then using continuity of the exponential function to interchange operators).

However I would like to share my incorrect attempt because I do not know exactly why it is incorrect. In particular there is one step to look at.

First I expanded the power $(1 + a_nx/n)^n$ out into a product and wrote "n times" below that. Then I used the fact that the limit of a product is a product of limits. However this step is not justifiable because how does one perform an infinite expansion?

(After that it's straightforward if one overlooks the above. For this product of limits, for each factor expand the limit of a sum into a sum of limits. So now you have an product of n sums: $lim 1 + lim a_nx/n$ n times. Finally you put the summand on the right as a product of limits so that this right summand becomes zero. Then you are left with a product of n limits $lim 1$ so that the answer is 1.)

$\endgroup$
  • $\begingroup$ Also, is there a way to prove this with the logarithm? I tried taking the logarithm on both sides for the equation that we want to arrive at. This yields $\lim_{n\to\ infty} \leftn ln (1 + a_nx/n)\right = ln 1 = 0$. This implies that the logarithm on the left hand side has to go to zero, correct? Because that's the only way to get zero on the right side. (I hope I'm correct in seeing that $n*0 = 0$ as n goes to infinity.) $\endgroup$ – David Apr 11 '15 at 15:52
5
$\begingroup$

You did this: $$\lim_{n\to \infty}\left(1+\frac{a_nx}{n}\right)^n=\lim_{n\to \infty}\underbrace{\left(1+\frac{a_nx}{n}\right)\left(1+\frac{a_nx}{n}\right)\cdots\left(1+\frac{a_nx}{n}\right)}_{n\text{ times}} $$

and because $\lim \frac{a_nx}{n}=0$ then the limit must be $1^n=1$. the mistake here is the fact that the number of terms of the product depends on $n$, if the terms of product does not depend on $n$ then you can use expansion on product.

In other words you did the following step: $$\lim_{n\to \infty}x_n^n=\lim_{n\to \infty}\underbrace{ x_nx_n\cdots x_n}_{n \text{ times}}=(\lim_{n\to \infty} x_n)^n$$

but as you know in the last term we have an $n$ (in the exponenet) outside of the limit and $n$ is defined only inside the limit

$\endgroup$
  • $\begingroup$ That was very well explained, +1. $\endgroup$ – Mattos Apr 11 '15 at 16:26
1
$\begingroup$

Using the reasoning in the argument above, you should get $$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=1 $$ because each of the $n$ terms of the product tend to $1$. However, this limit is $e=2.718281828\ldots$ This reasoning fails since $1^\infty$ is an indeterminate form, so you can't assume that just because each term of an infinite product goes to $1$ that the product is $1$.


Here is a proof using the inequality $1+x\le e^x$.

Since $\frac1{1+x}=1-\frac x{1+x}\le e^{-\large\frac x{1+x}}$, we get $$ e^{\large\frac x{1+x}}\le1+x\le e^x $$ Therefore, $$ e^{\frac{a_nx}{1+\frac{a_nx}n}}\le\left(1+\frac{a_n x}n\right)^n\le e^{a_nx} $$ By the Squeeze Theorem, we have $$ \lim_{n\to\infty}\left(1+\frac{a_n x}n\right)^n=1 $$

$\endgroup$
0
$\begingroup$

Probably the simplest way is to observe that $a_n \frac{x}{n} \to 0$ and hence

$$ \lim_n \left(1+a_n(x/n)\right)^{\frac{n}{xa_n}}=e$$

If you want to lose the logarithms, note that $$\lim_n \frac{\ln (1+a_n\frac{x}{n})-\ln1}{a_n\frac{x}{n}}=\ln'(1)=1$$

$\endgroup$
  • 1
    $\begingroup$ the question wasn't on how to prove it, the question was how to spot the error in a false proof – $\endgroup$ – Alan Apr 11 '15 at 16:06
  • 1
    $\begingroup$ @Alan The other answer addressed that already, but the OP added a comment to his question asking if it is possible to prove this by taking the log.... $\endgroup$ – N. S. Apr 11 '15 at 16:26
0
$\begingroup$

$\lim \space (1+a_n \frac x n)^n = \lim \space [(1+a_n \frac x n)^{\frac n {x a_n}}]^{a_n \frac x n} = \mathbb{e}^{\lim \space a_n \frac x n} = \mathbb{e}^0 =1$

$\endgroup$
  • 3
    $\begingroup$ the question wasn't on how to prove it, the question was how to spot the error in a false proof $\endgroup$ – Alan Apr 11 '15 at 16:06
  • $\begingroup$ @Alan: First, take a look at the title. Second, read the OP's own comment about how to prove this using the logarithm function. Third, deduce that in fact the OP wanted not only a verification of his proof (that he doubted), but also a certainly correct one. Fourth, cancel you downvote please. $\endgroup$ – Alex M. Apr 11 '15 at 17:20
  • $\begingroup$ Alan cannot take back a downvote as it is mine. Please read questions in full, not only the title. The first paragraph indicates there is no need just for some proof, as OP says the know a proof, and incidentally they describe this very proof. This answer seems thus not useful. (And, you do not use log either, which is why I dv'ed this question, yet not N.S. ones) $\endgroup$ – quid Apr 11 '15 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.