Sequences and Series

If ( d ) is a divisor of ( n ) that is relatively prime to ( n ), then ( \frac{n}{d} ) is also a divisor of ( n ) that is relatively prime to ( n ). Therefore, ( \mu(d) = \mu(\frac{n}{d}) ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \phi(\frac{n}{d}) ) for all ( d | n ) that are relatively prime to ( n ), we have ( \sum_{d|n} \mu(d) = \phi(n) ).

The generating function for the sequence {1, 1, 2, 3, 5, 8, 13, ...} is G(x) = 1 / (1 - x - x^2), which can be obtained using the formula for the generating function of a Fibonacci-like sequence.

The Fibonacci sequence is a sequence of numbers where each number is the sum of the two preceding numbers. The first two numbers in the sequence are 0 and 1. The next number is 0 + 1 = 1, the next number is 1 + 1 = 2, and so on. The 10th term in the Fibonacci sequence is 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144 = 55.

Gauss's formula states that the sum of the squares of the first n natural numbers is equal to n(n+1)(2n+1)/6.

Gauss's Formula states that the sum of the squares of the first (n) natural numbers is (\frac{n(n+1)(2n+1)}{6}).

The exponential generating function for the Stirling numbers of the second kind is (\frac{1}{(1-x)^n}), which can be derived using combinatorial arguments or by solving the recurrence relation for (S(n, k)).

According to Virasena, the formula for finding the sum of the first n natural numbers is $$S_n = \frac{n(n+1)}{2}$$

The equation (\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}) is known as the Basel problem, which was first solved by Leonhard Euler in the 18th century.

The series $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ is convergent because it is an alternating series that satisfies the alternating series test. The terms of the series decrease in absolute value, and the limit of the terms as $n\to\infty$ is 0. Therefore, the series converges.

The series $\sum_{n=1}^\infty \frac{1}{2^n}$ is a geometric series with first term $a = \frac{1}{2}$ and common ratio $r = \frac{1}{2}$. The sum of a geometric series is given by $S = \frac{a}{1-r}$. Substituting the values of $a$ and $r$, we get $S = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$. Therefore, the sum of the series is 1.

A series is absolutely convergent if the series of absolute values of its terms is convergent. The series $\sum_{n=1}^\infty \frac{1}{n^2}$ is absolutely convergent because the series $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent. The series $\sum_{n=1}^\infty \frac{n}{n+1}$ is not absolutely convergent because the series $\sum_{n=1}^\infty \frac{n}{n+1}$ is divergent.

The series $\sum_{n=1}^\infty \frac{n^2+1}{n^3+1}$ is convergent because it satisfies the limit comparison test. Comparing it to the series $\sum_{n=1}^\infty \frac{1}{n}$, which is convergent, we have $\lim_{n\to\infty} \frac{\frac{n^2+1}{n^3+1}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{n^3+n}{n^3+1} = 1$. Therefore, $\sum_{n=1}^\infty \frac{n^2+1}{n^3+1}$ is also convergent.

The series $\sum_{n=1}^\infty \frac{1}{n(n+2)}$ is a telescoping series. Rewriting the terms as $\frac{1}{n} - \frac{1}{n+2}$, we have $\sum_{n=1}^\infty \frac{1}{n(n+2)} = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+2}\right) = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots = \frac{1}{2}$.

The series $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ is divergent because it is an alternating series that does not satisfy the alternating series test. The terms of the series do not decrease in absolute value, and the limit of the terms as $n\to\infty$ is not 0. Therefore, the series diverges.

The series $\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$ is a telescoping series. Rewriting the terms as $\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$, we have $\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[\left(\frac{1}{1\cdot2} - \frac{1}{2\cdot3}\right) + \left(\frac{1}{2\cdot3} - \frac{1}{3\cdot4}\right) + \left(\frac{1}{3\cdot4} - \frac{1}{4\cdot5}\right) + \cdots\right] = \frac{1}{4}$.