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[tex]

\sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}

[/tex]

ln is not always bigger then 1

so when i am doing the comparing test

i cant use that

because ln 1 =0

??

\sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}

[/tex]

ln is not always bigger then 1

so when i am doing the comparing test

i cant use that

because ln 1 =0

??

Last edited: