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Because of the fact that Sum [(-1)^n*(n/ln(n)] is an alternating series, we'll solve it with Leibniz test.

The Leibniz test imposes 2 constraints:

1) Lt (un) = 0, n tends to inf.

2) u n+1 < u n

1) We'll note [n/ln(n)] = un

Lt un = Lt [n/ln(n)], when n tends to inf.

Lt [n/ln(n)] = inf./inf. (indetermination case).

We'll apply L'Hospital rule:

Lt [n/ln(n)] = Lt (n)'/[ln(n)]' = Lt 1/1/n = Lt n = inf.

2) We'll test the second condition, where u n+1=(n+1)/ln(n+1)

n+1>n

ln (n+1)>ln n

1/ln (n+1)<1/ln n

(n+1)/ln (n+1)< n/ln n

The second condition is accomplished.

**Because of the fact that the first condition for Leibniz test is not accomplished, because Lt (un)=0, when n tends to inf., the alternating series Sum [(-1)^n*(n/ln(n)] is divergent.**

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.

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