The Integral Test cannot be applied to the sum of a positive increasing function.

Author: paulvojta

OpenProblemLibrary/Indiana/Indiana_setSeries5IntegralTest/ur_sr_5_11.pg

@@ -144,7 +144,7 @@ $m = random(2,6,1);
 
 qa(~~@questions, ~~@answers,
 "\( \displaystyle \sum_{n=1}^\infty n e^{-$a n}  \)" , "CONV",
-"\( \displaystyle \sum_{n=1}^\infty n e^{$a n}  \)" , "DIV",
+"\( \displaystyle \sum_{n=1}^\infty n e^{$a n}  \)" , "NA",
 "\( \displaystyle \sum_{n=1}^\infty \frac{\ln{($d n)}}{n} \)" , "DIV",
 "\( \displaystyle \sum_{n=1}^\infty \frac{$b}{n \ln ($c n)} \)" , "DIV",
 "\( \displaystyle \sum_{n=1}^\infty \frac{$b}{n (\ln ($c n))^{$m}} \)" , "CONV",
@@ -213,18 +213,10 @@ EOT
 
 if ($slice[$i] == 1) {
 &SOLUTION(EV3(<<'EOT'));
-($j).  The function \(f(x) = x e^{$a x}\) is continuous and increasing because it is the product of 
-continuous increasing functions, so \(f(x) \geq f(1)=e^{$a}\) when \(1\leq x < \infty\).  Thus
-\[ \int_1^{\infty} f(x) \; dx \geq \int_1^{\infty} e^{$a} \; dx = \infty.\]  
-Since \(f\) is increasing it follows that 
-\[ f(n) \geq \int_{n-1}^n f(x)\; dx  \text{ for each } n=2,3,4,\cdots\]
-thus one can apply the integral test:
-\[ \begin{aligned} \sum_{n=1}^{\infty} f(n) &= f(1) + \sum_{n=2}^{\infty}f(n) \\ 
-   &\geq  f(1)+\sum_{n=2}^{\infty} \int_{n-1}^n f(x)\; dx  \\
-   &= f(1) + \int_1^{\infty} f(x)\; dx \\
-   & = \infty,
- \end{aligned}\]
-which shows that the series diverges.  
+($j).  The function \(f(x) = x e^{$a x}\) is increasing because it is
+the product of positive increasing functions.  Therefore it is not
+a decreasing function, so the Integral Test does not apply, and
+the correct answer is NA.
 
 EOT
 }