Calculus II: Convergence of Series with Positive Terms

[domabo]

Homework Statement

https://imgur.com/DUdOYjE
The problem (#58) and its solution are posted above.

Homework Equations

I understand that I can approach this two different ways. The first way being the way shown in the solution, and the second way, which my professor suggested, being a Direct Comparison Test.

Since I don't know how to write in Latex ( I apologize)... here's an image of relevant tests.
https://imgur.com/F2vgRiS
as well as information pertaining to the specifics of the problem: https://imgur.com/PmRdsEa

The Attempt at a Solution

I can see how the solution works besides the initial step. I do not know where the 1/192 comes from.

 

[member 587159]

The solution clearly refers to Exercise 31, which you didn't include (I think), so you might want to take a look there.

 

[domabo]

The solution clearly refers to Exercise 31, which you didn't include (I think), so you might want to take a look there.

I did include it under specifics pertaining to the problem

 

[Ray Vickson]

Homework Statement

https://imgur.com/DUdOYjE
The problem (#58) and its solution are posted above.

Homework Equations

I understand that I can approach this two different ways. The first way being the way shown in the solution, and the second way, which my professor suggested, being a Direct Comparison Test.

Since I don't know how to write in Latex ( I apologize)... here's an image of relevant tests.
https://imgur.com/F2vgRiS
as well as information pertaining to the specifics of the problem: https://imgur.com/PmRdsEa

The Attempt at a Solution

I can see how the solution works besides the initial step. I do not know where the 1/192 comes from.

Presumably you already know that $\ln n < n^q$ for any $q > 0$ and large enough $n$, so just choose $q$ to give $12 q - 9/8 <-1$, hence $q < 1/96.$ Any such $q$ will do, but $q =1/192$ gives a "nice" value to $12 q - 9/8.$