# $(a_n)$ is a monotone increasing sequence of integers. Prove that: $\lim_{n\to\infty}(1+\frac{1}{a_n})^{a_n}=e$

I'm asked to prove the question above.

I need to show that if $(a_n)$ is an increasing sequence of integers then: $\lim_{n\to\infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e$

I was thinking of showing that $\lim_{n\to\infty}(a_n)=\infty$ and then, by definition, I can say that there exists a natural number $N$, such that for every $n>N, a_n>0$ and so $(a_n)^\infty_{n=N}$ is a sub-sequence of $(1+\frac{1}{n})^n$ and therefore shares the same limit- $e$

Any ideas?

• $(a_n)$ is anyway a subsequence of the increasing sequence of integers. – астон вілла олоф мэллбэрг Apr 21 '16 at 5:48
• – u54112 Apr 21 '16 at 6:47
• @lastresort Indeed this is a duplicate but in the link there is no answer and that question is on hold, which is as good as the question not existing. – DonAntonio Apr 21 '16 at 7:19
• @Joanpemo Noted, will take more care next time. – u54112 Apr 21 '16 at 7:28
• your argument is correct and fully rigorous. – Paramanand Singh Apr 21 '16 at 9:40

Yes, your reasoning is correct. As there exists some $N$ for which $a_n>0$ for all $n\geq N$, it follows that $a_{N+k} \geq k$ (as $(a_n)$ is an increasing integer sequence).
Hence, $$\left(1+\frac{1}{n-N}\right)^{n-N} \leq \left(1+\frac{1}{a_n}\right)^{a_n} \leq e, \quad\text{for n > N}$$
Since $\;a_n\nearrow\infty$ , we get that $\;\left\{a_n\right\}\subset\Bbb N\;$ is a subsequence of $\;\{n\}_{n=1}^\infty=\Bbb N\;$ , and since the limit of any subsequence of a converging sequence exists and equals the limit of the sequence, we get what we want.