Finding if two sequences can be Big o and problem in definition of Big O

Question

I want to know if this is a correct assertion.

e$^n$/n$^5$=O(1/n)

What I did was,

n <= 4 ln(n) whenever n>=3
e^n  <= n^4
e^n <= n^5/n
e^n/n^5 <= 1/n
e^n/n^5 <= (1)(1/n)

Thus there exists c=1, and N=3 such that |e$^n$/n$^5$|<=c|1/n| whenever n>=N.Thus e$^n$/n$^5$=O(1/n)

Is this a correct approach?

In the definition given at our class it says that

Let {X$_n$} and {Y$_n$} be two different sequences.

If there are constants C and N such that |X$_n$|<=C{Y$_n$} whenever n>=N,then we say that X$_n$=O(Y$_n$)

I want to know of the constant C should be a positive constant or can it be any real number.And the N we select should it be an integer.And in the sequence {X$_n$} say {X$_n$}=n$^3$+n$^2$ are these n terms in sequence are natural numbers

$e^n$ grows soooooo much faster than $n^5$ as n approaches positive infinity — phoeagon, Nov 16 '13 at 07:39
n <= 4 ln(n) whenever n >= 3 That statement isn't true at all. — Dennis Meng, Nov 16 '13 at 08:26
How can I show that this is false That is it is not the case that e^n/n^5=O(1/n) — clarkson, Nov 16 '13 at 11:42

score 0 · Accepted Answer · answered Dec 19 '13 at 06:43

One way, not necessarily the best, is to use the series defining $e^n$: $$e^n = \sum_{k=0}^\infty \frac{n^k}{k!} \ge \frac{n^5}{5!}$$ (since all the terms are positive, the inequality holds when only one term is kept.) Hence, $$\frac{e^n}{n^5} \ge \frac{1}{5!}$$ for all $n$, contradicting the assertion that it is $O(1/n)$.

Finding if two sequences can be Big o and problem in definition of Big O

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