Solution of Problem 2

with b_n = 1/ln(n). We verify that the three conditions of the hold:

1 ln(n) ³ 0  for  all  n

1 ln(n+1) £ 1 ln(n)   for  all  n

(since ln(n) is positive and increasing for n ³ 2, its reciprocal is decreasing)

lim n® ¥  1 ln(n) = 0

(since

lim n® ¥  ln(n) = ¥

)

Consequently, the series

¥ å n = 2  an = ¥ å n = 2  (-1)n ln(n)

converges.

To determine whether the convergence is ,

we consider the series

¥ å n = 2  | an| = ¥ å n = 2  1 ln(n)

We can use either the or the ,

comparing our series to

¥ å n = 2  1 n

Using the , it follows that since [1/ln(n)] ³ [1/n] ³ 0, (which follows from n ³ ln(n))

the divergence of the harmonic series

¥ å n = 2  1 n

(i.e. a with p=1),

implies the divergence of

¥ å n = 2  1 ln(n)

and hence, conditional convergence of the original series.