Question

Determine whether the series is convergent or divergent.

\(\sum\limits_{{\rm{n = 1}}}^\infty {{{\left( {{\rm{ - 1}}} \right)}^{{\rm{n - 1}}}}\frac{{\sqrt {\rm{n}} }}{{{\rm{n + 1}}}}} \)

Short answer

The Alternating Series Test shows that the provided series is convergent.

Step-by-step solution

To use the Alternating Series Test.

Given:

\(\sum\limits_{{\rm{n = 1}}}^\infty {{{{\rm{( - 1)}}}^{{\rm{n - 1}}}}} \frac{{\sqrt {\rm{n}} }}{{{\rm{n + 1}}}}\)

The numerator's degree is smaller than the denominator's degree, therefore \(\frac{{\sqrt {\rm{n}} }}{{{\rm{n + 1}}}} \to {\rm{0}}\)is the same as\({\rm{n}} \to \infty \). The series converges using the Alternating Series Test because the terms are decreasing and the limit is zero.

Find out if the series is convergent or divergent.

It's getting smaller since the derivative of \({\rm{f(x) = }}\frac{{\sqrt {\rm{x}} }}{{{\rm{x + 1}}}}\)is negative for\({\rm{x > 1}}\).

\(\begin{array}{c}{{\rm{f}}'}{\rm{(x) = }}\frac{{\frac{{\rm{1}}}{{\rm{2}}}{{\rm{x}}{{\rm{ - 1/2}}}}{\rm{(x + 1) - }}\sqrt {\rm{x}} {\rm{(1)}}}}{{{{{\rm{(x + 1)}}}{\rm{2}}}}}{\rm{ \times }}\frac{{{\rm{2}}\sqrt {\rm{x}} }}{{{\rm{2}}\sqrt {\rm{x}} }}\\{\rm{ = }}\frac{{{\rm{(x + 1) - 2x}}}}{{{\rm{2}}\sqrt {\rm{x}} {\rm{(x + 1)}}}}\\{\rm{ = }}\frac{{{\rm{1 - x}}}}{{{\rm{2}}\sqrt {\rm{x}} {\rm{(x + 1)}}}}\end{array}\)

As a result, the Alternating Series Test shows that the provided series is convergent.