Evaluating ∑ln(n)/n: Bounded Values & n\geq3

[negation]

In determining whether ∑ln(n)/n converges or diverges, where n=[0,∞], ln(n)/n must be evaluated at the upper and lower bound. My notes breezed through this part without much explanation on finding the bounded values for ln(n)/n. I suspect it might have something to do with squeeze theorem.
Or rather, why is n \geq 3 a good value?
(I'm not sure if this should be homework because technically it isn't but some clarification on the question I have would be great!)

 

[pasmith]

In determining whether ∑ln(n)/n converges or diverges, where n=[0,∞], ln(n)/n must be evaluated at the upper and lower bound. My notes breezed through this part without much explanation on finding the bounded values for ln(n)/n. I suspect it might have something to do with squeeze theorem.
Or rather, why is n \geq 3 a good value?
(I'm not sure if this should be homework because technically it isn't but some clarification on the question I have would be great!)

Let N \in \mathbb{N}. If f : [N, \infty) \to \mathbb{R} is a strictly decreasing strictly positive function, then \sum_{n=N}^M f(n) can be bounded by <br /> \int_{N}^M f(x)\,dx \leq \sum_{n=N}^M f(n) \leq f(N) + \int_{N+1}^M f(x-1)\,dx <br /> = f(N) + \int_N^{M-1} f(x)\,dx. To see this, draw the graphs of f(x), f([x]) and f(x-1), where [x] is the greatest integer less than or equal to x and compare the areas under each. Thus by the squeeze theorem, <br /> \lim_{M \to \infty} \int_{N}^M f(x)\,dx \leq \lim_{M \to \infty}\sum_{n=N}^M f(n) \leq <br /> f(N) + \lim_{M \to \infty} \int_N^{M-1} f(x)\,dx \leq f(N) + \lim_{M \to \infty} \int_N^M f(x)\,dx so that the sum converges if and only if the integral converges.

\ln(x)/x is not strictly decreasing or strictly positive on [1,\infty) and isn't even defined for x = 0. It is strictly increasing on (0, e] and is then strictly decreasing and strictly positive on [e, \infty). Thus we must take N = 3 and split the sum as <br /> \lim_{M \to \infty} \sum_{n=1}^M \frac{\ln(n)}n = \frac{\ln(2)}2 + \lim_{M \to \infty} \sum_{n=3}^M\frac{\ln(n)}n.

 

[mathman]

ln(n)/n > 1/n (n>2). ∑1/n diverges, therefore ∑ln(n)/n diverges.

 

[negation]

Let N \in \mathbb{N}. If f : [N, \infty) \to \mathbb{R} is a strictly decreasing strictly positive function, then \sum_{n=N}^M f(n) can be bounded by <br /> \int_{N}^M f(x)\,dx \leq \sum_{n=N}^M f(n) \leq f(N) + \int_{N+1}^M f(x-1)\,dx <br /> = f(N) + \int_N^{M-1} f(x)\,dx. To see this, draw the graphs of f(x), f([x]) and f(x-1), where [x] is the greatest integer less than or equal to x and compare the areas under each. Thus by the squeeze theorem, <br /> \lim_{M \to \infty} \int_{N}^M f(x)\,dx \leq \lim_{M \to \infty}\sum_{n=N}^M f(n) \leq <br /> f(N) + \lim_{M \to \infty} \int_N^{M-1} f(x)\,dx \leq f(N) + \lim_{M \to \infty} \int_N^M f(x)\,dx so that the sum converges if and only if the integral converges.

\ln(x)/x is not strictly decreasing or strictly positive on [1,\infty) and isn't even defined for x = 0. It is strictly increasing on (0, e] and is then strictly decreasing and strictly positive on [e, \infty). Thus we must take N = 3 and split the sum as <br /> \lim_{M \to \infty} \sum_{n=1}^M \frac{\ln(n)}n = \frac{\ln(2)}2 + \lim_{M \to \infty} \sum_{n=3}^M\frac{\ln(n)}n.

Hi

Why isn't ln(x)/x strictly increasing on the domain [1,∞)?
ln(x)/x - ln(x-1)/(x-1) gives a positive increment and continues on as our domain tends towards infinity. Or is it categourically "not strictly decreasing" because it grows too slow?
To begin with, what is the definitition of strictly positive in this context?

 

[mathman]

ln(x)/x is decreasing -> 0.