# Help Proving that $\frac{(1+\frac{1}{t})^t}{e} = 1 -\frac{1}{2t} + O(\frac{1}{t^2})$ for $t\geq 1$

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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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? – esproff Jul 16 '12 at 6:13