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

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?

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    $\begingroup$ 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. $\endgroup$
    – Ian
    Nov 8 '18 at 21:57
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HINT

We have that

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

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

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  • $\begingroup$ How will that help him? $\endgroup$
    – Yanko
    Nov 8 '18 at 22:01
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    $\begingroup$ @Yanko See gimusi's answer $\endgroup$ Nov 8 '18 at 22:02
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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 $\lim_{n\rightarrow\infty}(0.9999 + \frac{1}{n})^n \leq \lim_{n\rightarrow\infty} (0.9999 + \frac{1}{10001})^n = 0$

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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.

What can you do instead:

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 $n\rightarrow 0$.

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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.

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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: $$\lim_{n\to\infty} a_n=\lim_{n\to\infty} (0.9999+\frac{1}{n})^n=0.$$

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