Does #sum_(n=1)^oo (ln n)/(n+2)# converge or diverge?

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Answer 1 · 2015-10-04T09:49:22

#sum_(n=1)^oo (ln n) / (n+2)# is divergent.

When #n > e#, #ln n > 1#.

So if #sum_(n=1)^oo 1/(n+2)# is divergent, then so is #sum_(n=1)^oo (ln n) / (n+2)#

#sum_(n=1)^oo 1/n = 1 + 1/2 + sum_(n=1)^oo 1/(n+2)#

So if #sum_(n=1)^oo 1/n# is divergent, then so is #sum_(n=1)^oo 1/(n+2)#

#sum_(n=1)^oo 1/n = sum_(k=0)^oo sum_(n=2^k)^(2^(k+1)-1) 1/n > sum_(k=0)^oo sum_(n=2^k)^(2^(k+1) - 1) 1/(2^k) = sum_(k=0)^oo 1# diverges.

That is:

#1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 +...#

#>= 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 +...#

#= 1 + 1/2 + 1/2 + 1/2 + ...#

#= oo#