SOLUTION: This is a calculus question. Can you explain each choice of #4 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200E3-F2018.pdf ?

Algebra ->  Equations -> SOLUTION: This is a calculus question. Can you explain each choice of #4 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200E3-F2018.pdf ?       Log On


   

Question 1195918: This is a calculus question. Can you explain each choice of #4 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200E3-F2018.pdf ?
Answer by ikleyn(52903) About Me  (Show Source):
You can put this solution on YOUR website!
.

(I) At small values of real positive quantity x, ln(1+x) is equivalent to x 
    as a small quantity in Calculus.

    Even more precisely,  0 < ln%281%2Bx%29 <= x

    THEREFORE,  0  <  sum%28ln%281%2B1%2Fn%5E2%29%2Cn=1%2Cinfinity%29 <= sum%28%281%2Fn%5E2%29%2Cn=1%2Cinfinity%29.

    The series  sum%28%281%2Fn%5E2%29%2Cn=1%2Cinfinity%29  is convergent, which is well known fact in Calculus after Euler. 

    So, the series  sum%28ln%281%2B1%2Fn%5E2%29%2Cn=1%2Cinfinity%29  is convergent, too.



(II) When x is small, sin(x) is equivalent to x as a small quantity in Calculus.

     Therefore, sum%28sin%281%2Fn%29%2C+n=1%2Cinfinity%29  is equivalent to sum%28%281%2Fn%29%2C+n=1%2Cinfinity%29.


     The series sum%28%281%2Fn%29%2C+n=1%2Cinfinity%29 is harmonic series, and it is divergent - which is

         a well known fact in Calculus.


     It implies that the series sum%28sin%281%2Fn%29%2C+n=1%2Cinfinity%29 is divergent, too.



(III) Estimate  sqrt%28n%5E3%29-sqrt%28n%5E3-1%29 = n%5E%283%2F2%29%2A%281-sqrt%281-1%2Fn%5E3%29%29  is equivalent to  n^(3/2)*(1/(2n^3)) = (1/2)*(1/n^(3/2)).

      Therefore, it is OBVIOUS that the series  sum%28%28sqrt%28n%5E3%29-sqrt%28n%5E3-1%29%29%2Cn=1%2Cinfinity%29  is convergent.


Thus we have this conclusion: series (I) converges; series (II) diverges; series (III) converges.


ANSWER.  Option (B).

Solved.


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Comment from student: Thank you. I am confused on the step that you explained  (I took a screenshot of it):
https://drive.google.com/file/d/1SMF_1rbZAQCY0SzU_9WfjtzSff8WLDbC/view?usp=sharing
How did you manipulate algebraically?


My response:   the question is to explain why

        1+-+sqrt%281-1%2Fn%5E3%29   is equivalent to   1%2F%282n%5E3%29.         (*)

Notice that in this problem  n  is a  BIG  value;  so,  1%2Fn  and  1%2Fn%5E3  are small quantities.

In  Calculus,  if  " a "  is a small value,  then

        sqrt%281%2Ba%29-1  is equivalent, in order of smallness,  to  a%2F2.         (**)

It is from the area of common knowledge in  Calculus.

So,  our estimate  (*)  is a direct consequense of  (**).


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If you seriously want to understand the solutions of this problem and other similar problems
from your list,  you should be  FREE  in manipulating basic knowledge of  Calculus,

since they are  PRE-REQUIZITES  for solving such problems.