# If $a_{n}\to+\infty$ then $(1+\frac{1}{a_{n}})^{a_{n}}\to e$

I already am given a solution to this that I don't quite understand. Here is how it goes:

If $n_{k}$ is any increasing sequence of positive integers then

$\lim_{k\rightarrow \infty } \ (1+\frac{1}{n_{k}+1})^{n_{k}}= \lim_{k\rightarrow \infty } \ (1+\frac{1}{n_{k}})^{n_{k}+1}=e$ [1]

The result follows from the following inequality using the Squeeze theorem and [1]: $(1+\frac{1}{\left \lfloor a_{n} \right \rfloor+1})^{\left \lfloor a_{n} \right \rfloor}\leq (1+\frac{1}{a_{n}})^{a_{n}}\leq (1+\frac{1}{\left \lfloor a_{n} \right \rfloor})^{\left \lfloor a_{n} \right \rfloor+1}$

I have two questions:

1. I don't know how to get that $\lim \ (1+\frac{1}{n_{k}+1})^{n_{k}}= \lim \ (1+\frac{1}{n_{k}})^{n_{k}+1}$

2. A sequence $a_{n}$ that goes $1,10,1,100,1,1000,...$ diverges to $+\infty$ but isn't increasing so I think we shouldn't be able to use [1]. Is this proof erroneous?

EDIT:

Thank you for all your insightful answers. You got me wondering now what if I had had: $\lim \ (\frac{1}{n_{k}+1})^{n_{k}}= \lim \ (\frac{1}{n_{k}})^{n_{k}+1}$

The suggested method for 1. wouldn't have worked because we get the indeterminate form $0.\infty$ so how could one do then without the squeeze theorem?

• For (2), $a_{n}$ diverges but not to $\infty$. – Seewoo Lee Jul 31 '16 at 20:35
• @John11 This may be a matter of definition, but I'd say the sequence in (2) does not diverge: its limit just doesn't exist when $\;n\to\infty\;$ . – DonAntonio Jul 31 '16 at 20:37
• @DonAntonio The very definition of divergent (for sequences) is not being convergent. – Kibble Jul 31 '16 at 20:43
• @DonAntonio How so? For any $M >0$ you take you can find a term in the sequence that is bigger. The following terms don't necessarily have to be all greater than $M$ do they? – John11 Jul 31 '16 at 20:46
• @John11 If the sequence converges in the wide sense of the word (sometimes also called "diverges to infinity"), then yes: if $\;M>0\;$ there exists $\;N\in\Bbb N\;$ such that $\;n>N\implies a_N>M\;$ . – DonAntonio Jul 31 '16 at 20:48

For question (1): Observe that $$\lim_{n\to\infty}\left(1+\frac{1}{n_{k}}\right)^{n_{k}+1}= \lim_{n\to\infty}\left(1+\frac{1}{n_{k}}\right)^{n_{k}}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n_{k}}\right)=e\cdot1=e,$$ and $$\lim_{n\to\infty}\left(1+\frac{1}{n_{k}+1}\right)^{n_{k}}=\frac{\lim_{n\to\infty}\left(1+\frac{1}{n_{k}+1}\right)^{n_{k}+1}}{\lim_{n\to\infty}\left(1+\frac{1}{n_{k}+1}\right)}=\frac{e}{1}=e.$$ Hence, the two limits are equal.

For question (2): We say that a sequence $\{a_n\}$ of real numbers diverges to $\infty$ if for any $M>0$ there is some $N\in\mathbb{N}$ such that for all $n\geq N$ we have $a_n>M$. Does your example satisfy this criteria?

In response to your edit: Observe that $$\left(\frac{1}{{n_k}}\right)^{n_k}> \left(\frac{1}{{n_k}+1}\right)^{n_k}>0\implies\lim_{n\to\infty}\left(\frac{1}{{n_k}+1}\right)^{n_k}=0,$$ and $$\left(\frac{1}{{n_k}}\right)^{n_k}>\left(\frac{1}{{n_k}}\right)^{n_k+1}>0\implies\lim_{n\to\infty}\left(\frac{1}{{n_k}}\right)^{n_k+1}=0,$$ both by the squeeze theorem.

• Thank you but (1) is not what I asked. I see that it is $e$ I'm confused about the indices flipping. – John11 Jul 31 '16 at 21:01
• @John11 My apologies I did not see the $+1$ in the denominator in the first question. Is that better? – Aweygan Jul 31 '16 at 21:10
• Yes Thank you! This has raised another question for me. Could you please check my edit in the question? – John11 Jul 31 '16 at 21:16
• @John11 Does that help? – Aweygan Jul 31 '16 at 21:27

Hint : $$(1+\frac{1}{n_k+1})^{n_k+1}$$ and $$(1+\frac{1}{n_k})^{n_k}$$

both tend to $e$, when $n_k$ tends to $\infty$. Furthermore , $1+\frac{1}{n_k}$ and $1+\frac{1}{n_k+1}$ both tend to $1$, when $n_k$ tends to $\infty$.

Take it from here.

1. The limit of $(1+1/n_k)$ is $1$, so by multiplying and dividing by it, doesn't change the limit, thus the result derives from the well known $(1+1/n)^n \to e$ and a variable change!

2. It doesn't diverge to infinity, its limit is undefined.