Comparison

Calculus Level 3

Let $a_n=\left( 1+\dfrac{1}{n} \right)^n$ , then which of the given options is true ?

a_{2007}>a_{2006}>a_{2008}

a_{2007}<a_{2006}<a_{2008}

a_{2006}<a_{2007}<a_{2008}

None of the given is true

a_{2006}>a_{2007}>a_{2008}

2 solutions

Curtis Clement
Feb 21, 2015

The definition of the exponential constant is given by: $\lim_{\ n\rightarrow \infty} (1+\frac{1}{n})^n = e\simeq 2.718$ . The function increases monotonically such that $a_n < a_{n+1}$ I'm a bit lazy, so a proof can be found here .

In other words, the given function is strictly increasing. I checked the link you gave and I think a simpler proof can be presented using derivatives. Taking $a_n=f(n)$ and finding $f'(n)$ will give you,

$f'(n)=f(n)\left\{ \ln\left(\frac{n+1}{n}\right)-\frac{1}{n+1}\right\}$

Clearly, $f'(n)\gt 0~\forall n\in \mathbb{N}$ . This verifies that $f(n)$ is strictly increasing. The "bounded above" condition can be showed by using the definition of $e$ as you did.

Prasun Biswas - 6 years, 3 months ago

It's not entirely obvious that $f'(n) \gt 0$ for all $n$ , since both terms inside the brackets go to $0$ as $n \rightarrow \infty.$ We could look at the Maclaurin series for $\ln(1 + x)$ for some guidance, as this would make the expression inside the brackets

$\ln(1 + \frac{1}{n}) - \frac{1}{n + 1} = \frac{1}{n} - \frac{1}{2n^{2}} + \frac{1}{3n^{3}} - \frac{1}{4n^{4}} + ..... - \frac{1}{n + 1}.$

which for "large enough" $n$ is essentially $\frac{1}{n} - \frac{1}{n + 1} \gt 0.$

P.S.. That wasn't my downvote, by the way. I don't know why anyone would downvote a perfectly valid comment.

Brian Charlesworth - 6 years, 3 months ago

This is an excellent explanation. The thing is, when I was writing that comment, I couldn't think of an elegant way to show that the derivative is $\gt 0$ . I verified it by deducing that $\ln\left(\dfrac{n+1}{n}\right)$ grows faster than $\dfrac{1}{n+1}$ for comparatively higher values of $n$ and the starting values of $f'(x)$ for natural $x$ were positive. That's why I used the word "clearly" to avoid writing more!

Prasun Biswas - 6 years, 3 months ago

@Prasun Biswas – Haha Yes, we know that $f'(x)$ is a bounded, increasing function, but those two terms are so close for large $n$ that it did require some proof. I used the term "for large enough $n$ " to avoid writing more myself. :)

Brian Charlesworth - 6 years, 3 months ago

@Brian Charlesworth – The curse of laziness strikes us both. Lol :P

Prasun Biswas - 6 years, 3 months ago

To show that $g(n)=\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}>0$ , let's find it's derivative: $g'(n)=\frac{1}{1+\frac{1}{n}}·\frac{-1}{n^2}+\frac{1}{(n+1)^2}=\frac{-1}{n(n+1)^2}<0$ . But $g(n)\rightarrow 0$ as $n\rightarrow\infty$ , so $g(n)$ decreases to 0, so it must be positive.

Laurent Shorts - 5 years, 2 months ago

Laurent Shorts
Apr 4, 2016

$a_n$ can be viewed as a 100% interest, but instead of giving 100% at the end of the interest period, we get $\frac{1}{n}$ of the interest at each $\frac{1}{n}$ of the period. The more often you get part of the interest, the more you'll get at the end of the period, as it is compound interest and not simple interest. That means $a_n$ increases.

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