why is the limit as n goes to infinity of $(1+\frac{1}{n}+\frac{200}{n^2})^n = e$?

Question

I know that $$\lim_{n\to\infty}\left(1+\frac{1}n\right)^n = e .$$

But why does $$\lim_{n\to\infty}\left(1+\frac{1}n+\frac{a}{n^b}\right)^n = e ? \quad where\quad b\gt1$$ better yet, how can I conclude something like: $$\lim_{n\to\infty}\left(1+\frac{1}n+\sum_{k=2}^\infty \frac{700^k}{k!n^k}\right)^n = e $$ Why do all the terms in the sigma not contribute anything to limit?

This is from a statistics course where we have to evaluate a similar expression but I have studied and done most of the exercises of the chapter on sequences and series of real numbers in Rudin's principles of math. analysis

Answer 1

HINT: $$ \left(1+\frac1n+\frac a{n^b}\right)^n=\left[\left(1+\frac{n^{b-1}+a}{n^b}\right)^{n^b}\right]^{1/n^{b-1}} $$

Answer 2

Compare each case to $e^{1+\epsilon} = \lim_{n \to \infty}(1 + \frac{1+\epsilon}n)^n$ and show that for any $\epsilon > 0$, there is an $N$ such that the expression in the above limit becomes bigger for any $n > N$. Therefore $e^1 = e$ is the largest number those could possibly converge to. And since they obviously do not converge to something less than $e$, you're done.

Answer 3

If $g(n)$ is a real valued function to the natural numbers with $n(g(n)-1)\to 0$ then $g(n)^n\to 1$. (Full proof on another question here.)

Then take $$g(n)=\frac{1+\frac1n + \frac a{n^b}}{1+\frac{1}{n}}\; b>1$$

Then we see that $$\left|n(g(n)-1)\right|=\dfrac{\dfrac {|a|}{n^{b-1}}}{1+\dfrac{1}{n}}<\dfrac{|a|}{n^{b-1}}\to 0.$$

So $g(n)^n \rightarrow 1$. That means that $\left(1+\frac{1}{n}+\frac{a}{n^b}\right)^n$ must converge to the same value as $\left(1+\frac{1}{n}\right)^n$, namely to $e$.

This works with all of your examples - take $g(n)$ as your expression divided by $1+\frac{1}{n}$.

More generally, if $h(n)=1+\frac{a}{n}+o\left(\frac{1}n\right)$, (assuming you know "little-$o$" notation[*]) then $$g(n)=\dfrac{h(n)}{1+\frac{a}{n}}$$ has the above property, and thus $h(n)^{n}\to e^{a}$.

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[*] The little-$o$ notation is equivalent of $n\left(h(n)-1-\frac{a}{n}\right)\to 0$.

Answer 4

$$ \left(1+\frac{1}{n}+\frac{a}{n^b}\right)^n=\left(1+\frac{n^{b-1}+a}{n^b}\right)^{\frac{n^b}{n^{b-1}+a} \frac{n^{b-1}+a}{n^b}\dot n}\to \exp\left({\lim\frac{n^{b}+an}{n^b}}\right)=e $$ If $b > 1$.

Answer 5

In situations like this loging helps. Write it out as $$ e^{n \log (1+ \frac{1}{n} + \frac{200}{n^2})} $$ You can take the limit 'inside' the exponential function because it is continuous. Now that you've done that, expand $\log( \cdot )$ into Taylor series around 1, which would work in this case becase the limit of the log function inside is $0$ at $n=\infty$: $$ e^{1 + \frac{1}{n} + \frac{200}{n}} \to e $$

Answer 6

I normally don't like using the $\log$ function when dealing with limits of this type, because there is a real danger of circularity in arguments. However, if we accept that $\lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e$, and, more generally, that $\lim_{n \to \infty}( 1 + \frac{x}{n})^{n} = e^{x}$ for any real number $x$, then we have an agreed starting point.

Now the inverse function to the exponential function is the natural logarithm function $\log x$ which has derivative $\frac{1}{x}.$ For any positive real number $x$, the mean value theorem gives $\frac{\log(1+x) - \log(1)}{x} = \frac{1}{1+\theta}$ for some $\theta \in (0,x)$, so that $\log(1+x) < x$.

Since it's more of a challenge, let's look at your second limit. I'll write it as $\lim_{n \to \infty}(1 + \frac{1}{n} + \sigma)^{n}$ for convenience. Now we have $\log( 1 + \frac{1}{n} + \sigma)^{n} = n \log( 1+ \frac{1}{n} + \sigma) \leq n ( \frac{1}{n} + \sigma ) \leq (1 + n \sigma)$.

It is clear that $\lim_{n \to \infty} n\sigma = 0$ from your formula for $\sigma$. Hence the log is tending to something at most $1$ as $n \to \infty$. However, we know that the limit itself is at least $e$, so the log we are taking is tending to $1$ as $n \to \infty$.

Since $\log$ is a continuous function, and the log of the expression is tending to $1$ as $n \to \infty$, the expression itself must have limit $e$.

Answer 7

Fix $a,b \in \mathbf{R}$ with $b>1$. Then for every $\varepsilon>0$ $$ \left(1+\frac{1-\varepsilon}{n}\right)^n \le \left(1+\frac1n+\frac a{n^b}\right)^n \le \left(1+\frac{1+\varepsilon}{n}\right)^n $$ whenever $n$ is sufficiently large. The claim follows taking the limit of each side.