# Finding $\lbrace a_{n}\rbrace$ s.t. $\mathop {\lim }\limits_{n \to \infty }a_{n}=1$ and $\mathop {\lim }\limits_{n \to \infty }a_{n}^{n}=2015$

The following problem appears in our analysis assignment.

Find a sequence $\lbrace a_{n}\rbrace$ of real numbers such that $$\mathop {\lim }\limits_{n \to \infty }a_{n}=1\text{ and }\mathop {\lim }\limits_{n \to \infty }a_{n}^{n}=2015.$$

Could anyone give me some help to find such a sequence ?

Any hints/ideas are much appreciated.

Thanks in advance for any replies.

• Cosider $a_n=2015^{1/n}$. – Hanul Jeon Feb 21 '15 at 14:33
• @tetori Oh its work! Thanks :) – ASB Feb 21 '15 at 14:36

$$a_n = 1+\frac{\log(2015)}{n}$$ deserves a try.