sequences and series HELP

[SCM4You]

Hello guys! As you can see, I am new here. I hope to be a contributing member one day, though!
I need some help with several problems related to sequences and series. If you are really good at them, please help me. I don't understand series at all. All of them are old FR questions(I think), with 2 MC questions.

1)The power series x+ ((x^2)/2) + ((x^3)/3) + ...+ ((x^n)/n) converges if and only if.
Answer from my teacher: -1 less than/equal to X < 1
I don't know why, though. What do I need to do to get the interval.


2) When we use (e^x) (approximate symbol)(1+x+ ((x^2)/2) to estimate the squareroot of e, the remainder is no greater than
a. 0.021 b. 0.034 c. 0.042 d. 0.067 e. 0.742
I'm stuck between b and d. Something is wrong with my steps here.


3) Let f be the function f(x)= 1/(x-1) and g be the function g(x)= (-1)/((x-1)^2)
a. find a power series, centered at o, for f
b. for what values of x does the series in part a converge?
c. Determine a power series for g
d. for what values of x does the series in part c converge?


4) Let f be the function f(x)= (x-5)/(xsquare -x-2)
a. write the function using partial functions
b. find a power series, centered at o, for f
c. for what values of x does the series in part b converge?



5) Let f be the function given by f(x)= e^(x/2)
a. write the first 4 nonzero terms and the general term for the Taylor series expansion of f(x) about x=0
b. use the result from part a to write the first 3 nonzero terms and the general term of the series expansion about x=0 for g(x)= ((e^(x/2))-1) / X

If you know how to do any of them, please help me.
Thank you.
~SCM4You.

 

[stapel]

Something is wrong with my steps...

Please reply showing your work and reasoning, so the tutors can help you find any errors.

Please be complete. Thank you.

Eliz.

 

[SCM4You]

Are the following answers right?

3.a.
f(x)=(1-x)^-1 f(0)=1
f?(x)=-(1-x)^-2 f(0)=1
f??(x)=2(1-x)^-3 f(0)=2
etc?
The series then becomes: 1+x+x2+ x3+ x4+ x5+ ?xn
3.b) no clue
3.c. ) Using the same idea as question 3a.
f(x)=-(x-1)-2 f(0)=-1
f?(x)=2(x-1)-3 f(0)=-2
f??(x)=-6(x-1)-4 f(0)=-(3!)
etc?
The Maclaurin is then written: (-1)-(2x)-(3x^2)-(4x^3)-?-nx^(n-1)
3.d. using the ratio test, i get -2<x<1. is this right?

4. a)
(x-5)/(x2-x-2) simplifies to (x-5)/[(x-2)(x+1)]. To write this in partial form:
(x-5)/[(x-2)(x+1)]=A/(x-2) +B/(x+1)
Cross multiplying we get:
x-5=A(x-1)+B(x-2).
When x=1, B=4
When x=2,A=-3.
We can then write the original function as: -3/(x-2)+4/(x-1).
4. b. We just write the power series for the above answer:
The power series for -3/(x-2) is : -1.5-(3x/4)-(3x^2/8)-(3x^3/16)?-3x^(n-1)/(2^n)
The power series for 4/(x-1) is: -4-(4x)-(4x^2)-(4x^3)?-4x^(n-1).


5.
f(x)= e^(0.5x) f(0)= 1
f?(x)=0.5 e^(0.5x) f(0)=0.5
f??(x)= 0.25e^(0.5x) f(0)=0.25
f???(x)= 1/8e^(0.5x) f(0)=1/8
The first 4 non-zero terms are: 1, 0.5, 0.25, and 1/8
The Maclaurin series is: 1+0.5x+(0.25x^2)/2+(x^3)/(8*3!)+?+(x^n)/[(2^n)*n!]
The general term is: (x^n)/[(2^n)*n!]

5. b.
((e^(x/2))-1)/ x =>[(1+0.5x+(0.25x^2)/2+(x^3)/(8*3!))-1]/x
This becomes:
[0.5x+(0.25x^2)/2+(x^3)/(8*3!)]x
The first three terms therefore are: 0.5+(0.25x)/2+(x^2)/(8*3!)
The general term is found by inspection and is: [(0.5^n)(x^(n-1))]/n!