Does L=∑∞i=21nlog(n) converge or diverge?
I established that:
L≤I=∫∞21nlog(n)=limn→∞[ln(log(n))−ln(log(2))],
and as limn→∞log(x)=∞, then L diverges.
But I'm not sure:
- of the sense of the inequality,
- about the conclusion.
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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