increasing, decreasing sequences or neither: 4 - ((-1)^n)/n,

[djdavis2k]

Determine whether the following sequences are (eventually) decreasing, eventually increasing, or neither. Explain

a.) {4- (((-1)^n)/n)}

b.) { ((n!)^2)/(2n)!}

c.) {(n^2*2^n)/(n!)}

d.) {(1.3.5...(2n-1))/((2n)^n)}

for a.) I believe it's neither increasing or decreasing because the (-1)^n/n is a harmonic sequence alternating between positive and negative values... however I do I write this explanation and method mathematically

for b.) I believe the sequence is increasing since (n!)^2/(2n)! can be compared to n^2/ (2n) =n/2 and assuming n is increasing the sequence should eventually increase... again.. i don't know if my explaination and mathematical proof is correct

for c.) I don't know I need help here

for d.) I believe the sequence is decreasing but I'm not sure need help here

 

[PAULK]

Determine whether the following sequences are (eventually) decreasing, eventually increasing, or neither. Explain

a.) {4- (((-1)^n)/n)}

b.) { ((n!)^2)/(2n)!}

c.) {(n^2*2^n)/(n!)}

d.) {(1.3.5...(2n-1))/((2n)^n)}

for a.) I believe it's neither increasing or decreasing because the (-1)^n/n is a harmonic sequence alternating between positive and negative values... however I do I write this explanation and method mathematically

>> Looks right to me. Suggestion for writing this 'rigorously' -- show that A[k+1] - A[k] is sometimes positive, and sometimes negative. If it were strictly increasing, that would always be positive, for example.

for b.) I believe the sequence is increasing since (n!)^2/(2n)! can be compared to n^2/ (2n) =n/2 and assuming n is increasing the sequence should eventually increase... again.. i don't know if my explaination and mathematical proof is correct
>> Looks wrong to me. But I suggest you look up Stirling's approximation to n! and apply it to this and the others.


for c.) I don't know I need help here

for d.) I believe the sequence is decreasing but I'm not sure need help here

>> You have 1.3.5...(2n-1), a product of odd integers. Let's insert the 'missing' even factors:

1.3.5...(2n-1)* 2.4.6...(2n)
------------------------------- =
2.4.6...(2n)

(2n)!
------
2^n (n!)

Maybe now you can do something with that.