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

Consider the series

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

If then the series diverges since as .

If then the series converges by the Leibniz rule since is decreasing and .

The convergence is absolute if and only if since

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

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

Consider the series

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 is a monotonic decreasing sequence with , then the alternating series converges. In this case we let

Then, is monotonic decreasing for . We can see this by considering

This is negative for ; hence, is decreasing for . Since is the sequence of the values of on the integers we have is decreasing if . Further, we have

Thus, we write,

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

Letting we have

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

diverges

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

Consider the series

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 is a monotonic decreasing sequence with , then the alternating series converges. In this case we have

is monotonic decreasing with . Hence, the alternating series

converges. This convergence is conditional since

diverges