Question
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
The correct answer is
P is divergent series; Q is convergent series
Let's examine the convergence or divergence of series P and Q step by step:
- Analysis of Series P: \(\sum_{n=1}^{\infty} \frac{1}{n}\)
- This series is the famous harmonic series.
- The general term of the series is \(\frac{1}{n}\).
- The harmonic series is known to be a divergent series.
- Reason: The partial sum of the harmonic series grows without bound as we add more terms. Formally, it can be shown using the integral test or by comparison with the integral \(\int_{1}^{\infty} \frac{1}{x} \, dx\), which diverges.
- Analysis of Series Q: \(\sum_{n=1}^{\infty} \frac{1}{n^2}\)
- This series is a p-series with \( p = 2 \).
- The general term of the series is \(\frac{1}{n^2}\).
- For a p-series \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), the series converges if \( p > 1 \).
- Since \( p = 2 \) in this series, it converges.
- Reason: The convergence of a p-series with \( p = 2 \) can be shown using the integral test or proven directly by comparing it to a converging integral, like \(\int_{1}^{\infty} \frac{1}{x^2} \, dx\), which converges.
Conclusion: Based on the explanations above, the correct option is that P is a divergent series; Q is a convergent series.
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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 infinite series:
$1+r+r^2 + r^3 + ............$ If
$r = 0.3$, then the sum of this infinite series is ________
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