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I'm trying to prove the asymptotic statement that for $t\geq 1$:

$$\frac{(1+\frac{1}{t})^t}{e} = 1 -\frac{1}{2t} + O(\frac{1}{t^2})$$

I know that $(1+\frac{1}{t})^t$ converges to $e$ and the right side looks like the first bit a of the series expansion for $e^{\frac{-1}{2t}}$ but I can't seem to work this thing out. Can someone point me in the right direction here, I'm new to asymptotic analysis and I don't really know how to deftly manipulate big-O notation yet. Thanks.

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2 Answers 2

up vote 3 down vote accepted

Here is what you do, replace $t$ by $\frac{1}{x}$ in your function, then compute Taylor Series at the point $x = 0$. This gives,

$$1-{\frac {1}{2}}x+{\frac {11}{24}}{x}^{2}-{\frac {7}{16}}{x}^{3}+{ \frac {2447}{5760}}{x}^{4}+O \left( {x}^{5} \right) $$

Now, substitute $x = \frac{1}{t}$ in the above series yields,

$$1-{\frac {1}{2t}}+{\frac {11}{24 t^2 }}-{\frac {7}{16 t^3}}+{ \frac {2447}{5760 t^4}} + O \left( {1/t}^{5} \right)\,.$$

For the big O notation, we say $f$ is a big O of $g$, if $|f(x)|\leq C|g(x)|$. Apply this definition to the above asymptotic series, you get the answer

$$ 1-{\frac {1}{2t}}+ O \left( {1/t}^{2} \right) \,.$$

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We prove by taking logarithm. Let $x = t^{-1}$. Then by the McLaurin expanstion of the logarithm,

$$\log \left[ \frac{\left(1 + \frac{1}{t}\right)^{t}}{e} \right] = \frac{\log(1 + x)}{x} - 1 = -\frac{x}{2} + O(x^2) = -\frac{1}{2t} + O\left(\frac{1}{t^2}\right).$$

Thus exponentiating,

$$ \frac{\left(1 + \frac{1}{t}\right)^{t}}{e} = \exp\left[ -\frac{1}{2t} + O\left(\frac{1}{t^2}\right) \right] = 1 -\frac{1}{2t} + O\left(\frac{1}{t^2}\right). $$

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Thanks for this, very slick, but could you please quickly explain how you derived the final equality? – Thoth Jul 16 '12 at 6:13
Cocktail M: Funny that you ask for an explanation about this (quite detailed) answer and accept another one (which completely omits the step you asked about) 30 minutes later. – Did Jul 16 '12 at 7:08
@did: could you explain the missing step for me? – Thoth Jul 16 '12 at 16:12
Cocktail: What for? You accepted the thing, this can only mean that you want to reward the poster for solving your problem, to inform others that your issue is resolved, and to indicate which answer you think is the most helpful to you. What could I add to this idyllic picture? – Did Jul 16 '12 at 19:13
Cock: I just reread the other post (the one you accepted): the really funny thing is that it does not even answer your question! No wonder you ask for more indications afterward... – Did Jul 20 '12 at 18:14

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