How do you use the direct comparison test to determine if #Sigma 1/(n-1)# from #[1,oo)# is convergent or divergent?

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Answer 1 · 2017-03-26T03:41:35

Note that #1/(n-1)>1/n#.

You can think about this logically by comparing that #1/2>1/3#, #1/9>1/10#, and so on.

Since #sum_(n=2)^oo1/n# is divergent #1/(n-1)>1/n#, we see that #sum_(n=2)^oo1/(n-1)# is divergent as well through the direct comparison test.

(Note that the question asks for #sum_(n=1)^oo1/(n-1)#, which is undefined for #n=1#, so I've begun the series at #n=2# instead.)