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Let $\{a_n\}$ be any sequence of real numbers such that $\lim_{n\rightarrow\infty}na_n=0$. Prove that $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$$

I thought about using binomial theorem. So $\left(1+\dfrac{1}{n}+a_n\right)^n = \left(1+\dfrac{1}{n}\right)^n + na_n\left(1+\dfrac{1}{n}\right)^{n-1} + \dbinom{n}{2}a_n^2\left(1+\dfrac{1}{n}\right)^{n-2}+\ldots$.

The first term has limit $e$, the second term has limit $0$ (because $na_n$ has limit $0$, and $\left(1+\dfrac{1}{n}\right)^{n-1} = \left(1+\dfrac{1}{n}\right)^{n}\left(1+\dfrac{1}{n}\right)^{-1}$ has limit $e$.) But for the other terms, it seems hard to find the limit.

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$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}+a_n\right)^n$$

$$= \left(\lim_{n\rightarrow\infty}\left(1+\frac{1+n\cdot a_n}n\right)^{\frac n{1+n\cdot a_n}}\right)^{\lim_{n\rightarrow\infty}(1+n\cdot a_n)}$$

$$= \left(\lim_{y\rightarrow\infty}\left(1+\frac1y\right)^y\right)^{\lim_{n\rightarrow\infty}(1+n\cdot a_n)}$$

$$=e^1$$

as $\lim_{n\rightarrow\infty} n\cdot a_n=0,n\rightarrow\infty \implies y=\frac n{1+n\cdot a_n}\to\infty$

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  • $\begingroup$ In fact, if $\lim_{n\rightarrow\infty} n\cdot a_n=m$ where $m$ is finite, the limit will be $e^{1+m}$ $\endgroup$ – lab bhattacharjee Jun 9 '13 at 4:34
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Here is a technique $$\left(1+\frac{1}{n}+a_n\right)^n= e^{n\ln\left(1+\frac{1}{n}+a_n\right)}= e^{n\left ( (\frac{1}{n}+a_n) -\frac{1}{2}(\frac{1}{n}+a_n)^2+\dots. \right) }=e^{\left( (1+na_n) -\frac{n}{2}(\frac{1}{n}+a_n)^2+\dots.\right)} $$

$$ = e^{\left( (1+na_n) -\frac{n}{2}\frac{(1+na_n)^2}{n^2}+\dots.\right)} .$$

Now, take the limit as $n\to \infty$ gives the limit $e$.

Note: We used the Taylor series of $\ln(1+x)$

$$ \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\dots. $$

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All the other terms $\binom {n} {k} a_{n}^{k} (1+\frac{1}{n})^{n-k}$ (where $k=1,2\dots$) contain the term "$na_{n}$", which we know tends to $0$ when $n\to\infty$. Hence, the rest of the terms are $0$.

For purposes of clarity, $\binom{n}{k}=\frac{n}{k}\binom {n-1}{k-1}$.

$na_{n}=0$.

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  • $\begingroup$ But maybe your "0" is multiplied by infinity. So I don't think you can conclude like that. $\endgroup$ – PJ Miller Jun 9 '13 at 3:58
  • $\begingroup$ You have stated in the question itself that $na_{n}=0$! Why would I conclude anything else? $\endgroup$ – fierydemon Jun 9 '13 at 4:01
  • $\begingroup$ No, I only stated that $\lim_{n\rightarrow\infty}na_n=0$, not $na_n=0$. $\endgroup$ – PJ Miller Jun 9 '13 at 4:03
  • $\begingroup$ Yes. But $n\to\infty$ even when we expand the binomial. If you look closely at the question, you'll realise it says $lim$ $n\to\infty$ $\endgroup$ – fierydemon Jun 9 '13 at 4:07
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For any $\epsilon>0$, choose an $N$ such that for all $n>N$, $|na_n|<\epsilon$.

Then the expression $$\frac{1}{e}\lim_{n\to\infty}\left(1+\frac{1}{n}+a_n\right)^n =\lim_{n\to\infty}\left(\frac{n+1+na_n}{n+1}\right)^n $$ and for all $n>N$, $$1-\frac{n\epsilon}{n+1}\leq\left(\frac{n+1-\epsilon}{n+1}\right)^n\leq\left|\left(\frac{n+1+na_n}{n+1}\right)^n \right|\leq \left(\frac{n+1+\epsilon}{n+1}\right)^n\leq \exp \left(\frac{n\epsilon}{n+1}\right)$$

Taking limits we obtain $$1-\epsilon \leq \left|\lim_{n\to\infty}\left(\frac{n+1+na_n}{n+1}\right)^n\right|\leq \exp(\epsilon)$$ But $\epsilon$ was arbitrarily small, so $$\left|\lim_{n\to\infty}\left(\frac{n+1+na_n}{n+1}\right)^n\right|=1$$ and the result follows$\square$

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  • $\begingroup$ Why do you say that $\left(1+\frac{1}{n}\right)^n \le \left(1+\frac{1}{n}+a_n\right)^n$? $\endgroup$ – wj32 Jun 9 '13 at 4:29
  • $\begingroup$ My mistake, I thought it was given that $a_n>0$. Let me rework my solution. $\endgroup$ – pre-kidney Jun 9 '13 at 4:31
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HINT:

What is $\frac{1}{n}+a_n$ equal to?

Proceed as you would if you were just finding $\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n$

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  • $\begingroup$ I don't get your hint what is $\frac{1}{n}+a_n$ equal to. What should it equal to? $\endgroup$ – PJ Miller Jun 9 '13 at 4:01
  • $\begingroup$ Just combining into one fraction, it's $\frac{1+n a_n}{n}$ $\endgroup$ – john Jun 9 '13 at 4:05
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This is easily solved using a lemma I learnt here on MSE:

Lemma: If $n(x_{n} - 1) \to 0$ as $n \to \infty$ then $x_{n}^{n} \to 1$ as $n \to \infty$.

We are already aware that $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ so to use the lemma we set $$x_{n} = \dfrac{1 + \dfrac{1}{n} + a_{n}}{1 + \dfrac{1}{n}}$$ so that $$n(x_{n} - 1) = \dfrac{na_{n}}{1 + \dfrac{1}{n}}\to \frac{0}{1 + 0} = 0$$ and thus $x_{n}^{n} \to 1$ which solves the problem.

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