Consider the two series,S_A and S_B, where  $S_A=\sum _{n=1}^{\infty }\frac{n^2}{2^n}$ and $S_B=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}+\frac{1}{128}+\frac{1}{512}+\dots$

Which of the following statements is correct for the two given series?

The correct statement is: Both S_A and S_B converge.

Let us analyze the convergence of the two series step-by-step.

1. Convergence of S_A:
The first series is defined as:
$$S_A=\sum _{n=1}^{\infty }\frac{n^2}{2^n}$$

We can determine the convergence of this series using the d'Alembert's Ratio Test. Let the general term of the series be:
$$a_n=\frac{n^2}{2^n}$$

The next term in the sequence, $a_{n+1}$, is:
$$a_{n+1}=\frac{{(n+1)}^2}{2^{n+1}}$$

Now, we calculate the limit of the absolute ratio of consecutive terms as n approaches infinity:
$$L=\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{lim}(\frac{{(n+1)}^2}{2^{n+1}}\times \frac{2^n}{n^2})$$

Simplifying the terms:
$$L=\underset{n\to \infty }{lim}(\frac{{(n+1)}^2}{n^2}\times \frac{2^n}{2^n·2})=\underset{n\to \infty }{lim}[{(1+\frac{1}{n})}^2\times \frac{1}{2}]$$

As n approaches infinity, $\frac{1}{n}\to 0$, so:
$$L={(1+0)}^2\times \frac{1}{2}=\frac{1}{2}$$

Since $L=\frac{1}{2}<1$, by the Ratio Test, the series S_A converges.

2. Convergence of S_B:
The second series is given as:
$$S_B=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}+\frac{1}{128}+\frac{1}{512}+\dots$$

Let us write the terms of this series as powers of 2 to observe the pattern:
$$S_B=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^9}+\dots$$

We can group this series into two separate infinite geometric series:
Group 1 (odd-positioned terms):
$$S_{B1}=1+\frac{1}{8}+\frac{1}{64}+\dots =\sum _{k=0}^{\infty }{\left(\frac{1}{8}\right)}^k$$
This is an infinite geometric series with first term a₁ = 1 and common ratio r₁ = 1/8. Since $\left|r_1\right|<1$, this sub-series converges.

Group 2 (even-positioned terms):
$$S_{B2}=\frac{1}{2}+\frac{1}{16}+\frac{1}{128}+\dots =\sum _{k=0}^{\infty }\frac{1}{2}{\left(\frac{1}{8}\right)}^k$$
This is also an infinite geometric series with first term a₂ = 1/2 and common ratio r₂ = 1/8. Since $\left|r_2\right|<1$, this sub-series converges.

Since both components of the sum are convergent geometric series, their sum S_B converges as well.
Specifically, we can compute the sum:
$$S_B=\frac{1}{1-\frac{1}{8}}+\frac{\frac{1}{2}}{1-\frac{1}{8}}=\frac{8}{7}+\frac{4}{7}=\frac{12}{7}$$

Therefore, both series S_A and S_B are convergent.