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.

  • 16
    $\begingroup$ Cosider $a_n=2015^{1/n}$. $\endgroup$ – Hanul Jeon Feb 21 '15 at 14:33
  • $\begingroup$ @tetori Oh its work! Thanks :) $\endgroup$ – ASB Feb 21 '15 at 14:36

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

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