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 $\frac{1}{2}$
Calculus Matching: Series, Limits & Fourier Coefficients
This solution provides step-by-step calculations to match entries from List-1 (P, Q, R) with appropriate values from List-2 (I, II, III).
P: Infinite Series Summation
We need to calculate the sum $ S = \sum_{n=1}^\infty \frac{1}{(n+2)(n+1)} $.
- Use partial fractions: $ \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2} $.
- Recognize the telescoping nature of the series: $ \sum_{n=1}^N \left( \frac{1}{n+1} - \frac{1}{n+2} \right) = \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{N+1} - \frac{1}{N+2} \right) = \frac{1}{2} - \frac{1}{N+2} $
- Find the limit as $N \to \infty$: $ S = \lim_{N \to \infty} \left( \frac{1}{2} - \frac{1}{N+2} \right) = \frac{1}{2} $
Result: P matches with III ($\frac{1}{2}$).
Q: Limit Evaluation using L'Hôpital's Rule
Evaluate the limit $ L = \lim_{x \to 0} \left( \frac{3}{x^2} \int_0^x \sin(t) dt \right) $.
- Integral evaluation: $ \int_0^x \sin(t) dt = [-\cos(t)]_0^x = 1 - \cos(x) $.
- Substitute into the limit: $ L = \lim_{x \to 0} \frac{3(1 - \cos x)}{x^2} $.
- This is the indeterminate form $\frac{0}{0}$. Apply L'Hôpital's Rule: $ L = \lim_{x \to 0} \frac{3 \sin x}{2x} $.
- The form is still $\frac{0}{0}$. Apply L'Hôpital's Rule again: $ L = \lim_{x \to 0} \frac{3 \cos x}{2} $.
- Evaluate the final limit: $ L = \frac{3 \cos(0)}{2} = \frac{3}{2} $.
Result: Q matches with I ($\frac{3}{2}$).
R: Fourier Series Coefficient Sum
For the function $f(x) = \frac{1}{2} \sin x - \frac{1}{2} \cos x + \frac{1}{\sqrt{2}} \sin 2x$, find the sum $ \sum_{n=0}^\infty (a_n^2 + b_n^2) $.
Comparing $f(x)$ to the standard Fourier series $f(x) \sim \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$, we identify the coefficients:
- $a_0 = 0$
- $a_1 = -\frac{1}{2}$
- $b_1 = \frac{1}{2}$
- $a_2 = 0$
- $b_2 = \frac{1}{\sqrt{2}}$
- All other coefficients ($a_n, b_n$ for $n \ge 3$) are 0.
Calculate the sum $ \sum_{n=0}^\infty (a_n^2 + b_n^2) = a_0^2 + \sum_{n=1}^\infty (a_n^2 + b_n^2) $:
$ = 0^2 + \left( a_1^2 + b_1^2 \right) + \left( a_2^2 + b_2^2 \right) + \sum_{n=3}^\infty (a_n^2 + b_n^2) $ $ = 0 + \left( \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 \right) + \left( 0^2 + \left(\frac{1}{\sqrt{2}}\right)^2 \right) + 0 $ $ = \left( \frac{1}{4} + \frac{1}{4} \right) + \left( 0 + \frac{1}{2} \right) = \frac{1}{2} + \frac{1}{2} = 1 $Result: R matches with II ($1$).
Final Matches Summary
The correct pairings derived from the calculations are:
- P $\longrightarrow$ III
- Q $\longrightarrow$ I
- R $\longrightarrow$ II
This corresponds to the option P $\longrightarrow$ III, Q $\longrightarrow$ I, R $\longrightarrow$ II.
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____.
