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.
Important Questions from Infinite Series
- Consider the two series, $S_A$ and $S_B$, where
$$S_A = \sum_{n=1}^\infty \frac{n^2}{2^n}$$
$$S_B = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \cdots$$
Which of the following statements is correct for the two given series? - The value of $\sum_{i=0}^{\infty} \sum_{j=1}^{\infty} 2^{-i} 3^{-j}$ is ______________ . (Answer in integer)
- Match each entry of List-1 with a suitable entry in List-2 and choose the correct option.
List-1 List-2 P The sum of the series $\sum_{n=1}^\infty \frac{1}{(n+2)(n+1)}$ is equal to I $\frac{3}{2}$ Q $\lim_{x \to 0} \left( \frac{3}{x^2} \int_0^x \sin(t) dt \right)$ is equal to II $1$ R Let $\frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$ be the Fourier series expansion of the function $f(x) = \frac{1}{2} \sin x - \frac{1}{2} \cos x + \frac{1}{\sqrt{2}} \sin 2x, x \in [0, 2\pi]$. Then, $\sum_{n=0}^\infty (a_n^2 + b_n^2)$ is equal to III $\frac{1}{2}$ The sum of the following infinite series is
$2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots$Consider the following two series
P: $\sum_{n=1}^{\infty} \frac{1}{n}$
Q: $\sum_{n=1}^{\infty} \frac{1}{n^2}$
Choose the correct option from the following
