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I am having difficulty finding a constant k such that k is larger than$ \frac{(logn)^b}{n} $ where b is a constant. Is there a way I can rewrite this?

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Since $\,\frac{\log^bn}{n}\xrightarrow[n\to\infty]{}0\,$ for any $\,b\,$ , do you want a bound for all $\,n\,$? – DonAntonio Feb 10 '13 at 20:52
Yes. I want a constant that is greater than the maximum of the function. For example for 1/n, the maximum is 1, so k would be 1 – phil12 Feb 10 '13 at 20:52
Is there an asymptotic maximum to this function? – phil12 Feb 10 '13 at 21:16
up vote 1 down vote accepted

noting the title and assuming that you want b>0 and n>1 and using ln instead of log

$$f'(n,b) = \frac{(b-ln(n))(ln ~n)^{(b-1)}}{n^2} = 0$$ $$b*(ln ~n)^{(b-1)} = (ln~ n)^b$$ $$n = e^b$$

$$ln(e^b)^b/(e^b) = b^b/(e^b)$$

$$k> \frac{b^b}{e^b}$$

does that work for you?

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Let me see if I understand your question correctly. You have $b$, a constant. You have the function $f(n) = \frac{\ln^b(n)}{n}$.

You wish you maximize the function $f(n)$ with respect to $n$, then find a $k$ that is larger than that maximum.

$f_n(n) = \frac{b\ln^{b-1}(n) - \ln^b(n)}{n^2} = 0$

$\therefore b = \ln(n) $

And so we know $n = e^b$ is the value that optimizes the function. Since there is only the one optimization point Im going to assume, based on the question, that this is a maximum. At the optimization point we have a value, a maximum, of $f(e^b) = (\frac{b}{e})^b$.

And value $k\gt (\frac{b}{e})^b$ will produce your result.

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