Here's the problem: Find the values of p for which the infinite sum of (ln k)/(k^p) converges.
I feel like I'm making it harder than it is. But it's looking for constants, and I've gotten p > [ln(k*lnk)]/(lnk) ... which is not a constant, and I think it definitely wrong. Any help is appreciated, thanks!
Charles L
Favorite Answer
Suppose r > 0
f(x) = ln x, g(x) = x^r
ln x/x^r = f(x)/g(x)
g'(x) = d/dx e^(r ln x) = r/x e^(r ln x) = (r/x)x^r
f'(x) = 1/x
f'(x)/g'(x) = (1/x)/[(r/x)x^r] = 1/(rx^r)
Therefore L'Hospital's Rule implies that:
lim f(x)/g(x) = 0
x →∞
Thus if x is large enough, then ln x/x^r = f(x)/g(x) < 1.
Suppose p > 1, then p = 1 + 2s for some s > 0, so
(ln k)/(k^p) = (ln k)/(k^s•k^(1+s)) = [(ln k)/(k^s)] k^(1+s)
∞
Σ 1/k^(1+s) <--- converges if p > 1
k=1
Therefore
∞
Σ (ln k)/(k^p) <--- converges if p > 1
k=1
Thanks, that was interesting!