The expressions above proves divergence by the nth term test, i.e. the limit of the sequence does not go to zero.
Here's a small survey of other techniques.
Is interesting to note that this sequence leads to an inconclusive Ratio Test, i.e. |T_n+1/T_n|=1.
Then 2*T_n = (2*n+4)/(2*n+1) = 1 + 3/(2*n+1).
So, T_n = (1/2) + 3/[2*(2*n+1)].
This expression has a somewhat easier limit to evaluate, however to get there required extra steps. From here though, its easier to approximate the sum.
The constant term, (1/2), guarantees each additional term in the sum adds a half, so part of the Partial Sum is (N/2).
The second term is (3/2)/(2*n+1)= (3/4)/(n+1/2)> (3/4)/(n+1),
so sum(T_n) > (N/2) + (3/4)*sum(1/(k+1), k, 1,N), with the right hand side being a lower bound for the sum.
The second sum is approximately the natural log of N.
So sum(T_n,1,n) > (N/2) + (3/4)*ln(N/e).