The Limit Of The Exponent

We have been given that $( 1+ \frac{1}{n})^{n+a_n} = e$ .

My first attempt is to solve for for $a_{n}$ .

Taking $ln$ on both sides,

$(n + a_n) * ln ( 1 +\frac{1}{n} ) = 1$

So, $(n+ a_n) = \frac{1}{ln(1+\frac{1}{n})}$

Thus , $a_n = \frac{1}{ ln (1+\frac{1}{n})} - n$

We have to find $\lim_{n \to \infty} \ a_n = ?$

i.e we will have to find

$\lim_{n \to \infty} \frac{1}{ ln (1 + \frac{1}{n})} - n$

We can treat it as a whole new problem. I prefer to change variable from $n \to \infty$ to $x \to 0$ by putting $x = \frac{1}{n}$

$\lim_{x \to 0^{+} } \frac{1}{ln(1+x)} - \frac{1}{x}$

$= \lim_{x \to 0^{+} } \frac{x - ln(1+x)}{x*ln(1+x)}$

Applying l'Hopital's rule ( i.e. differentiate the numerator and denominator w.r.t x ) as the expression is of the form $\frac{0}{0}$ ,

$\lim_{x \to 0^{+} } \frac{1 - \frac{1}{1+x}}{ln(1+x) + \frac{x}{1+x}}$

$= \lim_{x \to 0^{+} } \frac{1+x-1}{(1+x)*(ln(1+x) +x }$

$= \lim_{x \to 0^{+} } \frac{x}{(1+x)*(ln(1+x) +x }$

Again we see that at $x = 0$ the expression is indeterminate $(\frac{0}{0})$ .

On second application of l'Hopital's rule ,

$\lim_{x \to 0^{+} } \frac{1}{\frac{1}{1+x} + ln ( 1+x ) + \frac{x}{1+x} + 1}$

This expression at x = 0 gives value : $\frac{1}{1+0+0+1} = \frac{1}{2}.$

Therefore we get $\lim_{n \to \infty} \ a_n =$ $\boxed{\frac{1}{2}}$

Rearranging $\left(1+\frac{1}{n}\right)^{n+a_n}=e$ and taking limits gives $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left(\frac{1-n\log{\left(1+\frac{1}{n}\right)}}{\log\left(1+\frac{1}{n}\right)}\right)$

Using the Maclaurin series $\displaystyle\log(1+x)=\sum_{r\mathop=1}^{\infty}\frac{(-1)^{r+1}x^r}{r}$ it is easy to see the limit tends to $\boxed{\dfrac{1}{2}}$

We have $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n+a_n}=e$

Taking the $\log$ of both sides, we obtain $\lim_{n\to\infty} (n+a_n)\log\left(1+\frac{1}{n}\right)=1$

Note that the power series for $-\log(1-x)$ is $\sum_{k=1}^{\infty} \frac{x^k}{k}$ for $|x|<1$ . Therefore:

$\begin{aligned}\lim_{n\to\infty}(n+a_n)(n^{-1}-\frac12n^{-2}+\frac13n^{-3}+O(n^{-4}))&=1 \\ \lim_{n\to\infty}(n+a_n)(n^{-1}-\frac12n^{-2})&=1 \\ \lim_{n\to\infty}-\frac12n^{-1}+a_nn^{-1}-\frac12a_nn^{-2}&=0 \\ \lim_{n\to\infty}a_nn^{-1}&=\frac12n^{-1} \\ a_n&=\frac12\end{aligned}$

EDIT: attempt at fixing massive error in the proof:

Due to the power series of $\log(1-x)$ , we see that as $n$ gets arbitrarily large, $\log(1+n^{-1})$ gets arbitrarily close to $n^{-1}-\frac12n^{-2}$ . Let $a'_n$ be the solution to $(n+a'_n)(n^{-1}-\frac12n^2)=1$ . Then $a'_n=\frac1{\frac1n-\frac1{2n^2}}-n=\frac{1-(1-\frac12n^{-1})}{\frac1n-\frac1{2n^2}}=\frac12\frac{1}{1-\frac12n^{-1}}$

Note that $a_n=\frac1{\log(1-n^{-1})}-n$ , so $a_n\approx a'_n$ for arbitrarily large values of $n$

Since $\lim_{n\to\infty}a'_n=\lim_{n\to\infty}\frac12\frac{1}{1-\frac12n^{-1}}$ and $\lim_{n\to\infty}a_n=a'_n$ , $\lim_{n\to\infty}a_n=\frac12$

there is only one value between 0 & 1, that is 1/2 So, the solution is 1/2