The question asks to identify the statement that is NOT correct regarding the convergence of the power series $\Sigma_{n=1}^{\infty} \frac{(x+1)^n}{n \ 2^n}$.

Determine Radius of Convergence

We use the Ratio Test to find the radius of convergence ($R$). Let $a_n = \frac{(x+1)^n}{n \ 2^n}$.

$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(x+1)^{n+1}}{(n+1) \ 2^{n+1}} \cdot \frac{n \ 2^n}{(x+1)^n} \right| $

$ L = \lim_{n \to \infty} \left| \frac{x+1}{2} \cdot \frac{n}{n+1} \right| = \left| \frac{x+1}{2} \right| \lim_{n \to \infty} \frac{n}{n+1} = \left| \frac{x+1}{2} \right| \cdot 1 = \frac{|x+1|}{2} $

For convergence, $L < 1$, so $\frac{|x+1|}{2} < 1$, which implies $|x+1| < 2$. Thus, the radius of convergence is $R=2$. The series converges absolutely for $-1 < x < 1$. The interval of convergence is at least $(-1, 1)$.

Test Convergence at Endpoints

We need to check the convergence at the endpoints of the interval, which are $x = -1 - R = -3$ and $x = -1 + R = 1$. The center of the series is $x = -1$. The options involve checking specific values of $x$.

Convergence at $x = -3$

Substitute $x = -3$ into the series:

$ \Sigma_{n=1}^{\infty} \frac{(-3+1)^n}{n \ 2^n} = \Sigma_{n=1}^{\infty} \frac{(-2)^n}{n \ 2^n} = \Sigma_{n=1}^{\infty} \frac{(-1)^n}{n} $

This is the alternating harmonic series, which converges by the Alternating Series Test. Statement 1 is correct.

Convergence at $x = -1$

Substitute $x = -1$ into the series:

$ \Sigma_{n=1}^{\infty} \frac{(-1+1)^n}{n \ 2^n} = \Sigma_{n=1}^{\infty} \frac{0^n}{n \ 2^n} $

All terms are $0$ (for $n \geq 1$). The sum is $0$, which converges. Statement 2 is correct.

Convergence at $x = 0$

The value $x=0$ lies within the interval of absolute convergence $(-1, 1)$. Therefore, the series converges at $x=0$. Statement 3 is correct.

Explicitly:

$ \Sigma_{n=1}^{\infty} \frac{(0+1)^n}{n \ 2^n} = \Sigma_{n=1}^{\infty} \frac{1}{n \ 2^n} $

This series converges by the Ratio Test (as $L = 1/2 < 1$ calculated implicitly above).

Convergence at $x = 1$

Substitute $x = 1$ into the series:

$ \Sigma_{n=1}^{\infty} \frac{(1+1)^n}{n \ 2^n} = \Sigma_{n=1}^{\infty} \frac{2^n}{n \ 2^n} = \Sigma_{n=1}^{\infty} \frac{1}{n} $

This is the harmonic series, which is known to diverge. Statement 4 is NOT correct.

Conclusion

The statement that the series converges at $x = 1$ is incorrect.