Hello,

> Can sum(ln(n)/n^2 ~ 0.93 reach 1?

Consider the inequality

sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2, k=A..infinity)

The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2 yields

sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700

Clearly,

sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296

Summing these two inequalities shows that your sum is bounded above by

0.996. This estimate is quite rough, though; the actual value of the

infinite sum seems to be around 0.9375...

> Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?

Unfortunately, I don't know a nice trick for this one :-)

Peter

--

[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278