Question
Let $$ a_k = 2^{-k}k^4 \sin k \text{ and } b_k = 2^{-k^2} k \sin^2 k $$ for $ k = 1,2,.... $ Then
The correct answer is
both $ \sum_{k=1}^{\infty} a_k $ and $ \sum_{k=1}^{\infty} b_k $ converge
Series Convergence Analysis
We need to determine the convergence of two series: $ \sum_{k=1}^{\infty} a_k $ and $ \sum_{k=1}^{\infty} b_k $, where $ a_k = 2^{-k}k^4 \sin k $ and $ b_k = 2^{-k^2} k \sin^2 k $. We will analyze each series separately using the Absolute Convergence Test and the Comparison Test.
Convergence of $ \sum_{k=1}^{\infty} a_k $
- Consider the absolute value of the terms: $ |a_k| = |2^{-k}k^4 \sin k| = 2^{-k}k^4 |\sin k| $.
- Since $ |\sin k| \leq 1 $, we have $ |a_k| \leq 2^{-k}k^4 $.
- We test the convergence of $ \sum_{k=1}^{\infty} 2^{-k}k^4 $ using the Ratio Test. Let $ c_k = 2^{-k}k^4 $.
- Calculate the limit: $ \lim_{k \to \infty} \left| \frac{c_{k+1}}{c_k} \right| = \lim_{k \to \infty} \frac{2^{-(k+1)}(k+1)^4}{2^{-k}k^4} = \lim_{k \to \infty} \frac{1}{2} \left( \frac{k+1}{k} \right)^4 = \frac{1}{2} (1)^4 = \frac{1}{2} $
- Since the limit $ \frac{1}{2} < 1 $, the series $ \sum_{k=1}^{\infty} 2^{-k}k^4 $ converges by the Ratio Test.
- By the Comparison Test, since $ |a_k| \leq 2^{-k}k^4 $ and $ \sum_{k=1}^{\infty} 2^{-k}k^4 $ converges, the series $ \sum_{k=1}^{\infty} |a_k| $ converges.
- Therefore, $ \sum_{k=1}^{\infty} a_k $ converges absolutely and hence converges.
Convergence of $ \sum_{k=1}^{\infty} b_k $
- Consider the absolute value of the terms: $ |b_k| = |2^{-k^2} k \sin^2 k| = 2^{-k^2} k \sin^2 k $.
- Since $ \sin^2 k \leq 1 $, we have $ |b_k| \leq 2^{-k^2} k $.
- We test the convergence of $ \sum_{k=1}^{\infty} 2^{-k^2} k $. Let $ d_k = 2^{-k^2} k $.
- Calculate the limit using the Ratio Test: $ \lim_{k \to \infty} \left| \frac{d_{k+1}}{d_k} \right| = \lim_{k \to \infty} \frac{2^{-(k+1)^2} (k+1)}{2^{-k^2} k} = \lim_{k \to \infty} \frac{2^{-k^2-2k-1} (k+1)}{2^{-k^2} k} $ $ = \lim_{k \to \infty} \frac{1}{2} \frac{k+1}{k} 2^{-2k-1} = \lim_{k \to \infty} \frac{1}{4} \left( 1 + \frac{1}{k} \right) 2^{-2k} $
- As $ k \to \infty $, $ 2^{-2k} \to 0 $. Thus, the limit is $ \frac{1}{4} \times 1 \times 0 = 0 $.
- Since the limit $ 0 < 1 $, the series $ \sum_{k=1}^{\infty} 2^{-k^2} k $ converges by the Ratio Test.
- By the Comparison Test, since $ |b_k| \leq 2^{-k^2} k $ and $ \sum_{k=1}^{\infty} 2^{-k^2} k $ converges, the series $ \sum_{k=1}^{\infty} |b_k| $ converges.
- Therefore, $ \sum_{k=1}^{\infty} b_k $ converges absolutely and hence converges.
Conclusion
Both series $ \sum_{k=1}^{\infty} a_k $ and $ \sum_{k=1}^{\infty} b_k $ converge.
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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
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