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Let $\{a_{n}\}$ be any sequence of reals such that $\lim \limits_{n\rightarrow \infty } na_{n} =0$. Prove that $$\lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} + a_{n}\right)^{n} =e$$

I think I have shown that $\lim \limits_{ n\rightarrow \infty} \{ a_{n} \}$ must be zero and. I am thinking maybe showing the sequence above is bounded and decreasing might help. Basically I want to show it has the same limit as $\lim \limits_{n \rightarrow \infty } ( 1 + \frac{1}{n}) $ which is defined as e in my book.

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4 Answers

up vote 2 down vote accepted

Another approach:

If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ is $1^{+\infty}$, which is an indeterminate form, then: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$$

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@robjohn: Thanks for editing. Thanks for your support. –  B. S. Dec 24 '12 at 9:35
    
Well done again! :+) –  amWhy Mar 4 '13 at 0:42
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Hint: Rewrite the expression as $$\left(\left(1+\frac{1+na_n}{n}\right)^{\frac{n}{1+na_n}}\right)^{1+na_n}.$$

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Is this limit as form $1^{\infty}$? –  B. S. Dec 24 '12 at 7:43
    
Looks like it. I don't think of it quite that way, the inner expression has shape $(1+1/x)^x$ where $x$ is large. –  André Nicolas Dec 24 '12 at 7:48
    
May I ask to have a look at my answer if it is proper for this problem. I know this method from before. Thanks for your time. –  B. S. Dec 24 '12 at 7:50
    
@BabakSorouh: Looks correct. I had not seen the general assertion before. –  André Nicolas Dec 24 '12 at 8:04
    
Thanks for the time. –  B. S. Dec 24 '12 at 8:06
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One of the standard arguments goes as follows: Fix a sufficiently small $\varepsilon > 0$. Then by the assumption, there exists $N$ such that

$$ -\frac{\varepsilon}{n} \leq a_n \leq \frac{\varepsilon}{n} $$

for all $n \geq N$. Subsequently, we have

$$ \left(1 + \frac{1-\varepsilon}{n}\right)^{n} \leq \left(1 + \frac{1}{n} + a_{n}\right)^{n} \leq \left(1 + \frac{1+\varepsilon}{n}\right)^{n}. $$

In view of the limit

$$ \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n} = e^{x}, $$

we have

$$ e^{1-\epsilon} \leq \liminf_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} \leq \limsup_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} \leq e^{1+\epsilon}. $$

Since this is true for any small $\varepsilon > 0$. we conclude that

$$ \liminf_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} = \limsup_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} = e. $$

Therefore the convergence follows.

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First, let's factor out the part we know, then simplify $$ \begin{align} \lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} + a_{n}\right)^{n} &= \lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} \right)^{n} \left( \frac{1 + \frac{1}{n} + a_{n}}{1 + \frac{1}{n}} \right)^{n} \\ &= \lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} \right)^{n} \lim_{n \rightarrow \infty } \left( \frac{1 + \frac{1}{n} + a_{n}}{1 + \frac{1}{n}} \right)^{n} \\ &=e \lim_{n \rightarrow \infty } \left(1 + \frac{n a_{n}}{n + 1} \right)^{n} \end{align}$$

What's left looks sort of like the usual limit for $e$, so let's fiddle with it

$$ \begin{align} \cdots &=e \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n-1} \\ &= e \lim_{n \rightarrow \infty } \left(\left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{\frac{n-1}{n}} \\ &= e \left( \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{\lim_{n \rightarrow \infty } \frac{n-1}{n}} \\ &= e \left( \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{1} \\ &= e \lim_{n \rightarrow \infty } \left(1 + \frac{(n-1) a_{n-1}}{n} \right)^{n} \end{align} $$

Better. This looks even more like the limit for $e^x$; but the part that should be $x$ is going to $0$. So let's call it $L$ and bound it:

$$ \begin{align} L &\leq e \lim_{n \rightarrow \infty } \left(1 + \frac{x}{n} \right)^{n} \\ &= e^{1+x} \end{align} $$

$$ \begin{align} L &\geq e \lim_{n \rightarrow \infty } \left(1 - \frac{x}{n} \right)^{n} \\ &= e^{1-x} \end{align} $$

for all $x > 0$. That is, if $L$ is the limit, then

$$ e^{1-x} \leq L \leq e^{1+x} $$

Now we can use the method of exhaustion (i.e. squeeze theorem):

$$ e = \lim_{x \to 0^+} e^{1-x} \leq L \leq \lim_{x \to 0^+} e^{1+x} = e$$

and therefore $L = e$.

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Hi, when are you allowed to look at this as a limit raised to a limit as you did early on? –  Jmaff Dec 24 '12 at 22:22
    
Nice approach. + –  B. S. Dec 25 '12 at 11:04
    
@Jmaff: Whenever the limit form is one where exponentiation is continuous, such as when the base turns out positive and the exponent anything finite. You can reduce the question to that of a limit of a product and the logarithm of a limit by rewriting $$\lim x^y = \lim e^{y \log x} = e^{\lim y \log x} $$ –  Hurkyl Dec 25 '12 at 11:08
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