Question

Please Help!!!

Answer 1

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

$\dfrac12+\dfrac29+\dfrac4{27}+\dfrac8{81}+\cdots=\dfrac12+\dfrac{2^{2-1}}{3^2}+\dfrac{2^{3-1}}{3^3}+\dfrac{2^{4-1}}{3^4}+\cdots$

That is, we can write the sum more compactly as

$\dfrac12+\dfrac12\displaystyle\sum_{n=1}^\infty\left(\frac23\right)^n$

The series is geometric with common ratio $\dfrac23$, so the series converges (and thereby has a sum), so the answer is D.