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Determine the convergence of the series (-1)n-1 / ns

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

If s \leq 0 then the series diverges since a_n \neq 0 as n \to \infty.

If s > 0 then the series converges by the Leibniz rule since \frac{1}{n^s} is decreasing and \lim_{n \to \infty} \frac{1}{n^s} = 0.

The convergence is absolute if and only if s >1 since

    \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^{n-1}}{n^s} \right| = \sum_{n=1}^{\infty} \frac{1}{n^s} \]

converges if and only if s > 1 (by the integral test, Example #1 on page 398 of Apostol).

Determine the convergence of the series (-1)n (n1/2 / (n+100))

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} (-1)^n\frac{\sqrt{n}}{n+100}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The given series is conditionally convergent.

Proof. By the Leibniz rule we know that if \{ a_n \} is a monotonic decreasing sequence with \lim_{n \to \infty} a_n = 0, then the alternating series \sum_{n=1}^{\infty} (-1)^{n-1} a_n converges. In this case we let

    \[ a_n = \frac{\sqrt{n}}{n+100}. \]

Then, \{ a_n \} is monotonic decreasing for n > 100. We can see this by considering

    \[ f(x) = \frac{\sqrt{x}}{x+100} \quad \implies \quad f'(x) = \frac{100-x}{2\sqrt{x} (100+x)^2}. \]

This is negative for x > 100; hence, f(x) is decreasing for x > 100. Since \{ a_n \} is the sequence of the values of f(x) on the integers we have \{ a_n \} is decreasing if n> 100. Further, we have

    \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt{n}}{100+n} = 0. \]

Thus, we write,

    \begin{align*} \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n+100} &= \sum_{n=1}^{100} (-1)^n \frac{\sqrt{n}}{n+100} + \sum_{n=101}^{\infty} (-1)^n \frac{\sqrt{n}}{n+100}. \end{align*}

This converges since the first term is some finite number (since it is a finite sum) and the second term converges by the Leibniz rule; hence, the sum of these two terms converges.

Finally, the convergence is conditional since

    \[ \sum_{n=1}^{\infty} \left| (-1)^n \frac{\sqrt{n}}{n+100} \right| = \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n+100}. \]

Letting b_n = \frac{1}{\sqrt{n}} we have

    \[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{ \frac{\sqrt{n}}{n+100}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n}{n+100} = 1. \]

Hence, by the limit comparison test (and the fact that \sum b_n diverges) we conclude know that

    \[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n+100} \]

diverges. \qquad \blacksquare

Determine the convergence of the series (-1)n+1 / n1/2

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The given series is conditionally convergent.

Proof. By the Leibniz rule we know that if \{ a_n \} is a monotonic decreasing sequence with \lim_{n \to \infty} a_n = 0, then the alternating series \sum_{n=1}^{\infty} (-1)^{n-1} a_n converges. In this case we have

    \[ a_n = \frac{1}{\sqrt{n}} \]

is monotonic decreasing with \lim_{n \to \infty} a_n =0. Hence, the alternating series

    \[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}} \]

converges. This convergence is conditional since

    \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^{n+1}}{\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \]

diverges. \qquad \blacksquare