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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.

4 Answers 4

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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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    @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} $2012-12-25
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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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    @BabakSorouh: Looks correct. I had not seen the general assertion before.2012-12-24
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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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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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    Well done again! :+)2013-03-04