Test the series for convergence or divergence. sum[(-1)^n*(n/ln(n)) n=2..infinity] show steps

To test the convergence of (-1)^n*n/ln for n=2,3,....infinity,

Solution:

Sn = (2/ln2-3/ln3) + (4/ln4-5/ln5)+.....2n/ln2n-9n+1)ln(2n+1)+....

We study the difference (2n/ln - 2ln2n+1)

We know that (1+x)^n  >1+nx > nx .Or

n/ln(1+x) > ln(nx) . Or

n/ln(1+1)> lnn for x=1. Or

n/ln n >1/ln2. l/ln x is a continuous increasing function.

Therefore (n+1)/ln(n+1) - n/ln is posititive . So we can use cauchy's condensation test.

The Series Sn = Sum (2n+1)/ln(2n+1) -2n/ln(2n) and  sum Vn a^2n {a^(2n+1)/lna^(2n+1) - a^(2n)/lna^(2n)] where a is a a number >=2 behave alike.

Simplifying Vn:

Vn = a^(4n){a/[a/(2n+1)ln a] - 1/2nln a}

= (a^2n/lna){(a*2n-2n-1)/[(2n+1)(2n)]}

= (a^4n/lna){ 2n(a -1)+1]/[(2n)(2n+1)]}

= (a^4n/(2n+1){(a-1  +1/(2n)}{1/lna}

Taking limit a^4n/(2n+1) is unbounded. The other factor {a-1 +1/(2n){1/lna} is a finite quantity,

Therefore, Sum Vn diverges.And  Sn = Sum (2n+1)/ln(2n+1) -2n/ln(2n) should behave similarly.

Therefore , -Sn = Sum -[(2n+1)/ln(2n+1) -2n/ln(2n) ] = Sum{2n/ln2n - ( 2n+1)/ ln(2n+1)] should also diverge.