Let #s_n# be the #n#-th partial sum of the series, and consider the sums with index #n=3^k-1#:
#sigma_k = s_(3^n-1) = sum_(i=1)^(3^n-1) 1/(3i+1)#
Now we evaluate the difference between two consecutive such partial sums:
#sigma_(k+1) = sigma_k + 1/(3(3^n) +1) + 1/(3(3^n+1) +1) + ... + 1/(3(3^(n+1)-1) +1)#
We can see that all the terms that were added are larger than #1/3^(n+2)# and their number is:
#3^(n+1) - 1 -3^n +1 = 2*3^n#, so we have:
#sigma_(k+1) >= sigma_k + (2*3^n)*1 /3^(n+2)#
or:
#sigma_(k+1) >= sigma_k + 2/9#
Now if we start from:
# sigma_1 = s_2 = 1/4+1/7 = 11/28 > 2/9#
then we now that:
# sigma_2 >= sigma_1 +2/9 >2*2/9#
# sigma_3 >= sigma_2 +2/9 >3*2/9#
and in general:
# sigma_k > k*2/9#
so that we have:
#lim_(k->oo) sigma_k = oo#
and since the #sigma_k# are a subset of the partial sums of the series, then the series is proven to be divergent.
