Question

Determine whether the series is conditionally convergent, absolutely convergent, or divergent

\(\sum\limits_{{\rm{n = 2}}}^\infty {\frac{{{{{\rm{( - 1)}}}^{\rm{n}}}\sqrt {\rm{n}} }}{{{\rm{lnn}}}}} \)

Short answer

The series is divergent.

Step-by-step solution

Evaluate the relevant function's derivative.

Given: \(\sum\limits_{{\rm{n = 2}}}^\infty {\frac{{{{{\rm{( - 1)}}}^{\rm{n}}}\sqrt {\rm{n}} }}{{{\rm{lnn}}}}} \)

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

Check into it. \({{\rm{f}}'}{\rm{ = 0}}\) 

\(\begin{array}{c}{\rm{ln(x) - 2 = 0}}\\{\rm{lnx = 2x}}\\{\rm{ = }}{{\rm{e}}^{\rm{2}}} \approx {\rm{7}}{\rm{.4}}\end{array}\)

After the critical number, test the interval.

\(\left( {{{\rm{e}}{\rm{2}}}{\rm{,}}\infty } \right){\rm{:}}\;\;\;{{\rm{f}}'}{\rm{(8)}} \approx {\rm{0}}{\rm{.003 > 0}}\)

It is positive as\({\rm{x}} \to \infty \), so \({\rm{f}}\)is increasing for\({\rm{x > }}{{\rm{e}}{\rm{2}}}\).

Since \(\frac{{\sqrt {\rm{n}} }}{{{\rm{lnn}}}}\) is increasing

\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{{{\rm{( - 1)}}}{\rm{n}}}\sqrt {\rm{n}} }}{{{\rm{lnn}}}}{\rm{ = DNE}}\)

Because there is no limit, the series diverges using the Test for Divergence.

Therefore, the series is divergent.