$$S_A = \sum_{n=1}^\infty \frac{n^2}{2^n}$$
$$S_B = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \cdots$$
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
- Consider the following series:
(i) $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
(ii) $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$
(iii) $\sum_{n=1}^{\infty} \frac{1}{n!}$ - The sum of the following infinite series is:
$ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... $ - The series
$\sum_{n=0}^{r} q^n = 1 + q + q^2 + \dots$ has the sum: The value of the series $1+ \sin x + \cos^2 x + \sin^3 x + \dots$ at $x = \frac{ \pi}{4}$ is __________.
The sum of the infinite geometric series $1+\frac{1}{3}+\frac{1}{3^2} + \frac{1}{3^3} + ...$ (rounded off to one decimal place) is____.
