Does $\infty \sum n = 2 \frac{1}{n ln (n)}$ $sum_(n=2)^oo1/(nln(n))$ converges ?

Question

Does $\infty \sum n = 2 \frac{1}{n ln (n)}$ $sum_(n=2)^oo1/(nln(n))$ converges ?

1 Answer

Impact of this question

1099 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-06T02:58:52

The series diverges.

Explanation:

Use the integral test. If we let $u = ln n$ $u =lnn$ , then $d u = \frac{1}{n} d n$ $du = 1/ndn$ and $n d u = d n$ $ndu = dn$

$I = \int_{2}^{\infty} \frac{1}{n ln n} d n$ $I = int_2^oo 1/(nlnn) dn$

$I = \int_{2}^{\infty} \frac{1}{n u} \cdot n d u$ $I = int_2^oo 1/(n u) * ndu$

$I = \int_{2}^{\infty} \frac{1}{u} d u$ $I = int_2^oo 1/u du$

$I = {[ln u]}_{2}^{\infty}$ $I = [lnu]_2^oo$

$I = {[ln (ln n)]}_{2}^{\infty}$ $I = [ln(lnn)]_2^oo$

The limit $lim_(x->oo) ln(lnn)$ clearly equals $oo$ , therefore the series diverges.

Hopefully this helps!

Science

Math

Humanities

... and beyond

Does $\infty \sum n = 2 \frac{1}{n ln (n)}$ $sum_(n=2)^oo1/(nln(n))$ converges ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Does sum_(n=2)^oo1/(nln(n))∞∑n=21nln(n)sum_(n=2)^oo1/(nln(n)) converges ?

1 Answer

Explanation:

Related questions

Impact of this question

Does $\infty \sum n = 2 \frac{1}{n ln (n)}$ $sum_(n=2)^oo1/(nln(n))$ converges ?