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I'm taking a course on stochastic analysis. I'm stuck on the very first problem of the lecture notes:

$\lim_{n \to \infty} \left(1+\frac{\lambda}{n} + o(n^{-1})\right)^n = \exp(\lambda)$

Prior to the problem, the lecturer mentioned infinitesimal functions and introduced the Taylor series. I'm not sure how they are useful in proving the above, however.

I thought about taking the log of both sides:

$\lim_{n \to \infty} n\log \left(1+\frac{\lambda}{n} + o(n^{-1})\right) = \lambda$

then substituting $m=n^{-1}$ to get:

$\lim_{m \to 0} m^{-1} \log \left(1+\lambda m + o(m)\right) = \lambda$

I'm lost on where to go next. Does anybody have any ideas?

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As $m$ approaches 0, you can write $\log(1 + x)$ in terms of its taylor expansion. – TenaliRaman Dec 27 '12 at 11:04
Where does $x$ come from? – misha Dec 27 '12 at 11:25
$x$ in this case is $\lambda m + o(m)$. – TenaliRaman Dec 27 '12 at 11:29
up vote 1 down vote accepted

Write $$A = \lim_{n \to \infty} \left(1+\frac{\lambda}{n} + f(n)\right)^n$$ where $f(n) = o(n^{-1})$.

Then $$\log(A) = n\log\left( \lim_{n \to \infty} \left(1+\frac{\lambda}{n} + f(n)\right)\right)$$

Since $\lim_{n \to \infty} |f(n)| \leq \lim_{n \to \infty}\epsilon/n = 0$. It follows that

$$\log(A) = \log \left( \lim_{n \to \infty} \left(1+\frac{\lambda}{n}\right)^n\right)$$

and therefore:

$$ A = \exp(\lambda)$$

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Could you explain the reasoning behind the final step (the "therefore")? It's not obvious to me why $\lim_{n\to\infty}\left(1+\frac{\lambda}{n}\right)^n = \exp(\lambda)$. – misha Dec 27 '12 at 11:37
This holds by definition. – anthus Dec 27 '12 at 11:42
Ah, that's the part I was missing. Thank you. – misha Dec 27 '12 at 11:48

You can show and use the following inequality: $$\forall t\geqslant 0,\quad t-\frac{t^2}2\leqslant \log(1+t)\leqslant t.$$ It's classical and actually, it's a bound on the Taylor's remainder. But we can prove it directly.

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