B and C are absurd; if a series converges, it must have a sum, but if a series diverges, it cannot have a sum.
Now, notice that
[tex]\dfrac12+\dfrac29+\dfrac4{27}+\dfrac8{81}+\cdots=\dfrac12+\dfrac{2^{2-1}}{3^2}+\dfrac{2^{3-1}}{3^3}+\dfrac{2^{4-1}}{3^4}+\cdots[/tex]
That is, we can write the sum more compactly as
[tex]\dfrac12+\dfrac12\displaystyle\sum_{n=1}^\infty\left(\frac23\right)^n[/tex]
The series is geometric with common ratio [tex]\dfrac23<1[/tex], so the series converges (and thereby has a sum), so the answer is D.
