# Evaluate $\lim_{n\to \infty}(0.9999+\frac{1}{n})^n$

$$\lim_{n\to \infty}(0.9999+\frac{1}{n})^n$$

Using Binomial theorem:

$$(0.9999+\frac{1}{n})^n={n \choose 0}*0.9999^n+{n \choose 1}*0.9999^{n-1}*\frac{1}{n}+{n \choose 2}*0.9999^{n-2}*(\frac{1}{n})^2+...+{n \choose n-1}*0.9999*(\frac{1}{n})^{n-1}+{n \choose n}*(\frac{1}{n})^n=0.9999^n+0.9999^{n-1}+\frac{n-1}{2n}*0.9999^{n-2}+...+n*0.9999*(\frac{1}{n})^{n-1}+(\frac{1}{n})^n$$

A limit of each element presented above is 0. How should I prove that limit of "invisible" elements (I mean elements in "+..+") is also 0?

• An easier way to compute this limit is to instead assume $n>10000$ and then compare to $(0.9999+1/(10001))^n$ (or similar). Your way results in quite detailed asymptotics, which are nice if you're interested in large finite $n$ but not great for computing the limit.
– Ian
Nov 8 '18 at 21:57

HINT

We have that

$$\left(0.9999+\frac{1}{n}\right)^n=(0.9999)^n\left(1+\frac{\frac{1}{0.9999}}{n}\right)^n$$

Hint. Use that $$\lim (1 + \frac{x}{n})^n \to e^x$$ for any real number $$x.$$

• How will that help him? Nov 8 '18 at 22:01
• @Yanko See gimusi's answer Nov 8 '18 at 22:02

Hint Look at $$n\gt 10000$$

For $$n\gt 10000$$, we have $$0.9999 + \frac{1}{n} \leq 0.9999+ \frac{1}{10001}\lt1$$, so $$\lim_{n\rightarrow\infty}(0.9999 + \frac{1}{n})^n \leq \lim_{n\rightarrow\infty} (0.9999 + \frac{1}{10001})^n = 0$$

Wow your approach to this question is way too difficult.

Let me first explain what's wrong with your approach before I give you another way to do this.

You wrote your term as a sum of $$n$$ terms, each converges to zero as $$n$$ goes to infinity . In fact you didn't show that the "invisible" elements converges to zero, but even if you do, it is not enough. For instance the sum $$\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}$$ of $$n$$ times $$\frac{1}{n}$$ converges to $$1$$ even though each term converges to zero.

Instead you can say that for $$n$$ sufficiently large $$0.9999 + \frac{1}{n} < 0.99999$$ (add one more 9). Then use the fact that $$0.99999^n\rightarrow 0$$ as $$n\rightarrow 0$$.
Hint Instead of using the binomial expansion, observe that for $$n > 10^5$$ the quantity in parentheses is at most $$1 - \frac{1}{10^5 (10^5 + 1)}$$. Now apply the Squeeze Theorem.
Alternatively, consider the series: $$\sum_{n=1}^{\infty} (0.9999+\frac{1}{n})^n$$, which converges by the root test: $$\lim_{n\to \infty} a_n^{1/n}=\lim_{n\to \infty} (0.9999+\frac1n)=0.9999<1.$$ Hence: $$\lim_{n\to\infty} a_n=\lim_{n\to\infty} (0.9999+\frac{1}{n})^n=0.$$