Answer:
Step-by-step explanation:
a) 
Substitute limits to get
= 
Thus converges.
b) 10th partial sum =

=
c) Z [infinity] n+1 1 /x ^4 dx ≤ s − sn ≤ Z [infinity] n 1 /x^ 4 dx, (1)
where s is the sum of P[infinity] n=1 1/n4 and sn is the nth partial sum of P[infinity] n=1 1/n4 .
(question is not clear)
3 and four sevenths bcs 25divided by seven is 3r4
For this case we have the following inequality:

To solve we have:
Subtracting 6 from both sides of the equation:

Thus, the solution is given by all values of r greater than or equal to 5.
Answer:
The solution is given by all values of r greater than or equal to 5.
First pic: x=27
second pic: x=15
Answer:
c that is the answer yes .