Why does $\left(1+\frac1{x-0.5}\right)^x$ converge to $e$ quicker than $\left(1+\frac1x\right)^x$ does?

Question

As something adjacent to the birthday problem, I found that $\frac1{\ln{365}-\ln{364}}$ is very nearly $364$ or $365$, but much closer to $364.5$. This leads to $$\ln{\frac{x+0.5}{x-0.5}}\approx \frac1x.$$ Rearrangement leads to $$\left(1+\frac1{x-0.5}\right)^x\approx e$$ especially as $x$ tends to infinity. This resembles the limit $\lim_{x\to \infty}\left(1+\frac1x\right)^x=e$, but graphical analysis shows this tends to $e$ much slower than the other function. Why is this true?

Answer 1

Let $f(x) = \left(1+\frac{1}{x}\right)^x$, then

\begin{align} \ln f(x) &= x \ln\left(1 + \frac{1}{x}\right) \\ &= x \left( \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \dots \right) \\ &= 1 - \frac{1}{2x} + \frac{1}{3x^2} - \dots \end{align}

so that

$$f(x) = e^{1 - \frac{1}{2x} + \cdots} \approx e \cdot e^{-\frac{1}{2x}} \approx e \left(1 - \frac{1}{2x}\right)$$

but now consider $g(x) = \left(1+\frac{1}{x-0.5}\right)^x$. To clean up $g$, we can let $y = x - 0.5$ so that $$g(y) = \left(1 + \frac{1}{y}\right)^{y+0.5}$$

By the same process as above, we have that

\begin{align} \ln g(y) &= (y + 0.5) \left( \frac{1}{y} - \frac{1}{2y^2} + \frac{1}{3y^3} - \dots \right) \\ &= \left[ 1 - \frac{1}{2y} + \frac{1}{3y^2} \dots \right] + \left[ \frac{1}{2y} - \frac{1}{4y^2} + \dots \right] \\ &= 1 + \frac{1}{12y^2} - \dots \end{align}

so that

$$g(x) \approx e \left(1 + \frac{1}{12x^2}\right)$$

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Thus, $g$ converges faster than $f$, by a comparison of the error terms. By centering $f$, to get $g$, the $y^{-1}$ terms cancel out, leaving you with a better error term.

Answer 2

$$f=\left(1+\frac{1}{x+a}\right)^x$$

As other answerers, taking logarithms, using Taylor series and exponentiating again $$f=e + e \sum_{n=1}^\infty (-1)^n\frac {\alpha_n}{\beta_n\,x^n}$$ where the $\beta_n$ correspond to sequence $A053657$ in $OEIS$ and the first $\alpha_n$ are $$\small\left( \begin{array}{cc} n & \alpha_n \\ 1 & \color{red}{2 a+1} \\ 2 & 36 a^2+36 a+11 \\ 3 & \color{red}{(2 a+1)} \left(52 a^2+52 a+21\right) \\ 4 & 17520 a^4+35040 a^3+31320 a^2+13800 a+2447 \\ 5 & \color{red}{(2 a+1)} \left(24048 a^4+48096 a^3+46952 a^2+22904 a+4795\right) \\ 6 & 16333632 a^6+49000896 a^5+71784720 a^4+61901280 a^3+32180652 a^2+9396828 a+1190215 \\ \end{array} \right)$$

So, if $a=-\frac 12$ all odd terms disappear, and, as other wrote $$f=e + e \Bigg(\frac{1}{12 x^2}+\frac{23}{1440 x^4}+ \frac{1223}{362880 x^6}+O\left(\frac{1}{x^8}\right) \Bigg)$$

Answer 3

If you ask Alpha to expand near infinity, you get $$\left(1+\frac1{x-0.5}\right)^x\approx e + \frac e{12 x^2} + \frac{23 e}{1440 x^4} + O((1/x)^5)$$ which is missing the linear error term from

$$\left(1+\frac1x\right)^x \approx e - \frac e{2 x} + \frac {11 e}{24 x^2} - \frac{7 e}{16 x^3} + \frac {2447 e}{5760 x^4} + O((1/x)^5)$$ Your correction has eliminated the linear error term.

Answer 4

Recall the Taylor expansions $$\ln(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}k\quad\text{and}\quad\ln(1-x)=-\sum_{k=1}^\infty\frac{x^k}k,$$ which may be combined as $$\ln\frac{1+x}{1-x}=2x\sum_{k=0}^\infty\frac{x^{2k}}{2k+1}\quad(|x|<1).$$ Notice how, for small $x$, the latter series converges faster than the original two, because the powers of $x$ skip ahead in steps of two. This desirable feature is captured in your formula (pardon the change of variable) $$\mathrm e=\lim_{t\to\infty}\left(1+\frac1{t-\tfrac12}\right)^{\!t}.$$ This can be seen if we take the logarithm and use the substitution $t:=1/2x$. Thus: $$1=\lim_{x\to0}\left(\frac1{2x}\ln\frac{1+x}{1-x}\right).$$ We may therefore write this as $$\mathrm e=\lim_{x\to0}\exp\left(\frac1{2x}\ln\frac{1+x}{1-x}\right)=\lim_{x\to0}\exp\sum_{k=0}^\infty\frac{x^{2k}}{2k+1}.$$ Here, if we expand the exponential expression, only even powers of $x$ will occur.

Answer 5

Another solution $$f=\left(1+\frac{1}{x+a}\right)^x$$ $$1+\frac{1}{x+a}=\frac {1+t}{1-t} \quad \implies \quad \log(f)=\frac{1-(2 a+1) t}{2 t}\,\log \left(\frac{1+t}{1-t}\right)$$

At this point,we already see that, letting $a=-\frac 12$ will remove all odd powers of $t$ in the expansion

Expand as series around $t=0$ using $$\log \left(\frac{1+t}{1-t}\right)=2\sum_{n=0}^\infty \frac{t^{2 n+1}}{2 n+1} $$ So $$\log(f)=1+\sum_{n=0}^\infty \alpha_n\,t^n$$ with $$\alpha_{2n}=\frac{1}{2 n+1}\qquad \text{and} \qquad \alpha_{2n+1}=-\frac{2 a+1}{2 n+1}$$ So, if $a=-\frac 12$ all odd powers disappear and, back to definition, $t=\frac 1{2x}$ and $$\log(f)=1+\sum_{n=0}^\infty \frac 1{(2n+1)\,(2x)^{2n}}=1+\frac{1}{12 x^2}+\frac{1}{80 x^4}+\frac{1}{448 x^6}+O\left(\frac{1}{x^{8}}\right)$$ Exponentiating $$f=e+e\,\Bigg(\frac{1}{12 x^2}+\frac{23}{1440 x^4}+\frac{1223}{362880 x^6}+O\left(\frac{1}{x^{8}}\right) \Bigg)$$