- #1
Marcivo
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Hello, folks. This happens to be my first post here, and I've come with a question from a problem set in my textbook.
Determine whether the following series converges or diverges. Give reasons for your answer.
[tex]\sum^{\infty}_{n=2} \frac{1}{ln(n)^2}[/tex]
Now, I've taken a number of approaches toward this problem, but, so far, none of them has provided even an incorrect solution. The integral test fails automatically because the expression is nonintegrable; both the ratio and root tests fail because the solution is equal to 1; I've compared it to [tex]\frac{1}{ln(n)}[/tex], and [tex]\frac{1}{n^2}[/tex], as my b_n's, but both are inconclusive by the limit comparison test. The one thing I can think of that would prove the series to be divergent is to say that 1/ln(x)^2 is greater than 1/x, but is this a safe assumption to make?
I went to my professor today, but was just turned away after being told to "apply integration by parts" to the integral of 1/ln(x)^2...which, of course, doesn't work. So, my question is, does anyone here suggest a more solid comparison than 1/x, and, if not, how could I prove that 1/x [tex]\leq[/tex] 1/ln(x)^p for all p [tex]\geq[/tex] 2?
Thanks.
Homework Statement
Determine whether the following series converges or diverges. Give reasons for your answer.
Homework Equations
[tex]\sum^{\infty}_{n=2} \frac{1}{ln(n)^2}[/tex]
The Attempt at a Solution
Now, I've taken a number of approaches toward this problem, but, so far, none of them has provided even an incorrect solution. The integral test fails automatically because the expression is nonintegrable; both the ratio and root tests fail because the solution is equal to 1; I've compared it to [tex]\frac{1}{ln(n)}[/tex], and [tex]\frac{1}{n^2}[/tex], as my b_n's, but both are inconclusive by the limit comparison test. The one thing I can think of that would prove the series to be divergent is to say that 1/ln(x)^2 is greater than 1/x, but is this a safe assumption to make?
I went to my professor today, but was just turned away after being told to "apply integration by parts" to the integral of 1/ln(x)^2...which, of course, doesn't work. So, my question is, does anyone here suggest a more solid comparison than 1/x, and, if not, how could I prove that 1/x [tex]\leq[/tex] 1/ln(x)^p for all p [tex]\geq[/tex] 2?
Thanks.