Needs the Series converges, and its sum

`sum _(n=2)^oo ln n/n`

To determine if the series converges or diverges, apply integral test.

`int_2^oo lnn/n dn`

`= lim_(t->oo) int_2^t ln n/n dn`

To take the integral, apply u-substitution method.

`u=ln u`

`du = 1/n dn`

`int lnn/n = int ln n * 1/ndn = int u du = u^2 /2= (ln n)^2/2`

Then, plug-in the upper and lower limit to evaluate the improper integral.

= `lim_(t->oo) [ (ln n)^2/2]|_2^t`

= `lim_(t->oo) [(lnt)^2/2- (ln2)^2/2]`

And, take the limit of it as t approaches infinity.

`= oo`

Since `int_2^oo lnn/n dn` is divergent, therefore the series `sum_(n=2)^oo ln/n` is divergent.

Comments

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