How to prove divergence of infinite series 1/(n(sqrt(ln(n))))

what theory should i use? and also are the partial sums of this series unbounded? if yes find a partial sum guaranteed to be greater than 100... Im horrible at series please help!

First of all, start this series at n = 2, since 1/(n(sqrt(ln(n)))) is undefined at n = 1 since ln(1) = 0.

Use the integral test. Note that the terms are positive and decreasing, so this test can be used.

integral 2 to infinity of 1/(x(sqrt(ln(x)))) dx
= integral ln(2) to ln(infinity) of 1/(sqrt(u)) du, using u = ln x, du = (1/x)dx
= integral ln(2) to infinity of 1/(sqrt(u)) du
= 2 sqrt(u) evaluated from u = ln(2) to u = infinity
= infinity - 2 sqrt(ln(2))
= infinity

So the series diverges, and because all the terms are positive, the partial sums are unbounded.

From drawing a picture of over-approximating rectangles of width 1 onto the region above the x-axis under the curve y = 1/(x(sqrt(ln(x)))) from x = 2 to x = k + 1, we see that

sum from n = 2 to k of 1/(n(sqrt(ln(n))))
>= integral 2 to (k + 1) of 1/(x(sqrt(ln(x)))) dx
= 2 sqrt(u) evaluated from u = ln(2) to u = ln(k + 1) essentially from our work before
= 2[sqrt(ln(k + 1)) - sqrt(ln(2))]

Note that 2[sqrt(ln(k + 1)) - sqrt(ln(2))] > 100 is equivalent to

sqrt(ln(k + 1)) - sqrt(ln(2)) > 50
ln(k + 1) > [50 + sqrt(ln(2))]^2
k > exp{[50 + sqrt(ln(2))]^2} - 1

So the partial sum from n = 2 to k of 1/(n(sqrt(ln(n)))) is guaranteed to exceed 100 if
k > exp{[50 + sqrt(ln(2))]^2} - 1.