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The question is, find the limit of $\ a_n = (e^n+3^n)/5^n$ as $\ n→∞ $ I tried using L'hopital's rule, but didn't seem to get anything useful, so I figured I may be able to use the squeeze theorem. Would this be an appropriate use of the squeeze theorem, and would there be a better method to proving that the limit approaches zero?

$\ 1/n ≤ (e^n+3^n)/5^n ≤ 10/n $ for all n, with both $\ 1/n $ and $\ 10/n $ approaching zero as n approaches infinity.

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  • $\begingroup$ Yes, that is fine, if you can actually prove the two bounds. $\endgroup$ – Igor Rivin Dec 10 '13 at 3:39
  • $\begingroup$ Do you know the property, given $a \in \mathbb{R}$, such that $|a| < 1$, then $\lim a^n = 0$? $\endgroup$ – user49685 Dec 10 '13 at 3:43
  • $\begingroup$ Ok, that property combined with the answers given, prove that the limity approaches zero. $\endgroup$ – Dan Minor Dec 10 '13 at 3:47
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Hint: I don't believe the bounds you have are correct. Instead, I suggest something like $$0\leq\frac{e^n+3^n}{5^n}\leq\frac{3^n+3^n}{5^n}=2\cdot\frac{3^n}{5^n}.$$ These inequalities are much more obvious, hold for all $n$, thus we can take the limit. What do we get?

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It might be easiest to think of it as $\displaystyle a_n = \Big(\frac{e}{5}\Big)^n + \Big(\frac35\Big)^n $.

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