Would it be possible determining which elements in $(0.9999+\frac{1}{n})^n$ are the largest and smallest?
I know that $$\left( 0.9999 + \frac1n \right)^n = \sum_{k=0}^{n}\binom{n}{k} 0.9999^k \cdot \left( \frac1n \right)^{n-k}$$
I want to use in in order to find the limit of $(0.9999+\frac{1}{n})^n$ by applying the sandwich rule.
To find the largest summand in the expansion of $(a+b)^n$ (where $a,b>0$), note that $$\frac{ {n\choose k+1}a^{k+1}b^{n-k-1}}{{n\choose k}a^kb^{n-k}}=\frac{k!(n-k)!a}{(k+1)!(n-k-1)!b}=\frac{(n-k)a}{(k+1)b}.$$ Thus the $k$th summand is maximal when this quotient becomes $\le 1$ for the frst time. In your case, for whic $k$ does $\frac{n(n-k)0.9999}{k+1}\le 1$? You will find that this happens when $$ k\approx \frac{0.9999n^2-1}{1+0.9999 n}\approx n+1.$$ Conclude.