$$S_A = \sum_{n=1}^\infty \frac{n^2}{2^n}$$
$$S_B = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \cdots$$
Which of the following statements is correct for the two given series?
Series Convergence Analysis
We need to determine the convergence of the two given infinite series, $S_A$ and $S_B$.
$S_A$ Convergence Test
The first series is given by $S_A = \sum_{n=1}^\infty \frac{n^2}{2^n}$. We can use the Ratio Test to check for convergence.
Let $a_n = \frac{n^2}{2^n}$. Then $a_{n+1} = \frac{(n+1)^2}{2^{n+1}}$.
Applying the Ratio Test:
$ L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{(n+1)^2}{2^{n+1}} \cdot \frac{2^n}{n^2} \right| $ $ L = \lim_{n\to\infty} \left( \frac{n+1}{n} \right)^2 \cdot \frac{2^n}{2^{n+1}} $ $ L = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^2 \cdot \frac{1}{2} $ $ L = (1)^2 \cdot \frac{1}{2} = \frac{1}{2} $Since $L = \frac{1}{2} < 1$, the series $S_A$ converges by the Ratio Test.
$S_B$ Convergence Analysis
The second series is $S_B = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \cdots$.
Let's examine the terms and their denominators:
- Term 1: $1 = \frac{1}{2^0}$
- Term 2: $\frac{1}{2} = \frac{1}{2^1}$
- Term 3: $\frac{1}{8} = \frac{1}{2^3}$
- Term 4: $\frac{1}{16} = \frac{1}{2^4}$
- Term 5: $\frac{1}{64} = \frac{1}{2^6}$
- Term 6: $\frac{1}{128} = \frac{1}{2^7}$
- Term 7: $\frac{1}{512} = \frac{1}{2^9}$
The exponents in the denominators are $0, 1, 3, 4, 6, 7, 9, \dots$. This sequence consists of all non-negative integers $n$ such that $n$ is not congruent to $2$ modulo $3$ (i.e., $n \not\equiv 2 \pmod 3$).
Therefore, we can express $S_B$ as:
$ S_B = \sum_{n=0, n\not\equiv 2 \pmod 3}^\infty \frac{1}{2^n} $This sum can be found by taking the sum of all powers of $1/2$ and subtracting the sum of the missing powers ($n \equiv 2 \pmod 3$):
$ S_B = \left( \sum_{n=0}^\infty \frac{1}{2^n} \right) - \left( \sum_{k=0}^\infty \frac{1}{2^{3k+2}} \right) $The first part, $\sum_{n=0}^\infty (\frac{1}{2})^n$, is a geometric series with first term $a=1$ and common ratio $r=\frac{1}{2}$. It converges to $\frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.
The second part, $\sum_{k=0}^\infty \frac{1}{2^{3k+2}} = \frac{1}{2^2} + \frac{1}{2^5} + \frac{1}{2^8} + \cdots$, is a geometric series with first term $a = \frac{1}{2^2} = \frac{1}{4}$ and common ratio $r = \frac{1}{2^3} = \frac{1}{8}$. It converges to $\frac{a}{1-r} = \frac{1/4}{1 - 1/8} = \frac{1/4}{7/8} = \frac{1}{4} \times \frac{8}{7} = \frac{2}{7}$.
Thus, $S_B = 2 - \frac{2}{7} = \frac{14-2}{7} = \frac{12}{7}$. Since $S_B$ sums to a finite value, it converges.
Conclusion
Both series $S_A$ and $S_B$ converge.
The correct statement is: Both $S_A$ and $S_B$ converge.
Important Questions from Infinite 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 followingConsider the following infinite series:
$1+r+r^2 + r^3 + ............$ If
$r = 0.3$, then the sum of this infinite series is ________
