3
$\begingroup$

I understand that $$ \lim_{n\rightarrow \infty} \left(1 + \frac{1}{n}\right)^n = e $$

However, how would $ \lim_{n\rightarrow \infty} \left(1 + \frac{1}{3n}\right)^{5n} $ be simplified.

The $5$ can be moved outside the limit:

$$ {\lim_{n\rightarrow \infty} \left[\left(1 + \frac{1}{3n}\right)^{n}\right]^5} $$

But how can you simplify it to the form $e^{\frac{m}{n}}$?

The answer is $ e^{\frac{5}{3}} $, but could someone help me understand the methodology?

$\endgroup$
1
  • 6
    $\begingroup$ $\left\{\left(1+\frac{1}{3n}\right)^{3n}\right\}_{n\geq 1}$ is a subsequence of $\left\{\left(1+\frac{1}{n}\right)^{n}\right\}_{n\geq 1}$, hence it converges to the same limit, $e$. Given the continuity of $x\mapsto x^{5/3}$ over $\mathbb{R}^+$, your limit equals $e^{5/3}$. $\endgroup$ Sep 3, 2018 at 19:05

5 Answers 5

13
$\begingroup$

Don't move $5$. Move $5/3$ instead: $$\lim_{n\rightarrow \infty} (1 + \frac{1}{3n})^{5n}=\left(\lim_{n\rightarrow \infty} (1 + \frac{1}{3n})^{3n}\right)^{5/3}.$$ You will find the inside limit is $e$.

$\endgroup$
6
$\begingroup$

$\lim_{n\rightarrow \infty} (1 + \frac{1}{3n})^{5n} =\lim_{n\rightarrow \infty} ((1 + \frac{1}{3n})^{3n})^{5/3}= e^{5/3}$

$\endgroup$
3
$\begingroup$

Everybody has given the solution, but this is how it hit me,

$$\lim_{n\to \infty}\biggr(\bigg(1+\frac{1}{3n}\bigg)^n\biggr)^5$$

Let $$3n=t$$

$$\lim_{t\to \infty}\biggr(\bigg(1+\frac{1}{t}\bigg)^{\frac{t}{3}}\biggr)^5$$

$$\lim_{t\to \infty}\biggr(\bigg(1+\frac{1}{t}\bigg)^t\biggr)^\frac{5}{3}$$

$$e^\frac{5}{3}$$

$\endgroup$
1
$\begingroup$

As an alternative

$$\left(1 + \frac{1}{3n}\right)^{5n}=e^{5n\log \left(1 + \frac{1}{3n}\right)}\to e^{5/3}$$

indeed

$$5n\log \left(1 + \frac{1}{3n}\right)=\frac53\cdot\frac{\log \left(1 + \frac{1}{3n}\right)}{\frac1{3n}}\to\frac53 \cdot 1 = \frac53$$

$\endgroup$
1
$\begingroup$

Consider the most general case of $$y=\left(1 + \frac{1}{an}\right)^{bn}$$ Take logarithms $$\log(y)=bn\log\left(1 + \frac{1}{an}\right)$$ Now, use Taylor series $$\log(y)=bn\left(\frac{1}{a n}-\frac{1}{2 a^2 n^2}+O\left(\frac{1}{n^3}\right) \right)=\frac{b}{a}-\frac{b}{2 a^2 n}+O\left(\frac{1}{n^2}\right)$$ Use again Taylor $$y=e^{\log(y)}=e^{\frac{b}{a}}-\frac{b e^{\frac{b}{a}}}{2 a^2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .