- #1
Lo.Lee.Ta.
- 217
- 0
Sum from 0 to infinity of [3/[(n+1)(5^n)]](x-3)^n HELP :)
1.
∞
Ʃ [itex]\frac{3}{(n+1)(5^{n}}[/itex]*(x-3)n
n=0
2. The first question I had to answer was: What is f(3)?
I found the first 4 terms to be:
3, 3/10(x-3), 3/75(x-3)2, 3/189(x-3)3
So f(3) equals 3, I'm pretty sure.
Because all the other (except the 1st) cancel to equal 0.
3 + infinity zeros = 3
The second question asks me whether or not f(4) converges.
Well, that would be: 3 + (3/10)(1)1 + (3/75)(1)2+(3/189)(1)3 + ...
3 + increasingly small numbers = convergence
Is this the right way to think about it? Is this is correct, is there a better, perhaps faster way?
The third question asks whether or not f(8) converges.
That would be: 3 + (3/10)(5)1 + (3/75)(5)2 + (3/189)(5)3 + ...
= 3 + increasingly small numbers = convergence
...I don't think it's getting bigger very fast... It seems like it would take way too long for this to go to infinity. That's why I'm thinking it would converge...
The fourth question asks whether or not f(-2) converges.
That would be: 3 + (3/10)(5)1 + (3/75)(5)2 + ...
Yeah, this is the same as last time. So I say it converges again.
Now, the fifth question asks what the interval of convergence is for this power series.
To find this interval, I think I need to use the formula: an/an+1
[itex]\frac{3(x-3)^{n}}{(n+1)5^{n}}[/itex]*[itex]\frac{(n+2)(5^{n+1})}{3(x-3)^{n+1}}[/itex]
= [itex]\frac{5(n+2)}{(n+1)(x-3)}[/itex]
Now, the next step is to take the limit, but in this form, I'll get ∞/∞.
I could do L'Hospital's Rule, but I think it might be easier if I rewrote the problem before I take the limit.
Hopefully, I'm not doing any illegal math moves... :/
[itex]\frac{5}{(n+2)^{-1}(n+1)(x-3)}[/itex]
=[itex]\frac{5}{n^{0}+n^{-1}+2n^{-1}+2{-1}(x-3)}[/itex]
= [itex]\frac{5}{1 + 1/n + 1/2n + 1/2(x-3)}[/itex]
lim |[itex]\frac{5}{1.5 + 1/n + 1/2n(x-3)}[/itex]|
n→∞
= |5/[(1.5 + 0 + 0)(x-3)]|
I CAN'T FIND THE GREATER THAN SYMBOL, so I'm using "∠"!
= -1 ∠ 10/3(x-3) ∠ 1
= -33/10 ∠ x ∠ 33/10
...So it is convergent between this interval... Which I'm pretty sure is WRONG!
That is too hideous of an answer to be correct. AND- it does not fit with my convergence answers from before!
33/10= 3.3
4 AND 8 lie outside of this interval. Something's wrong here... :/
Please help! D=
Thanks :)
1.
∞
Ʃ [itex]\frac{3}{(n+1)(5^{n}}[/itex]*(x-3)n
n=0
2. The first question I had to answer was: What is f(3)?
I found the first 4 terms to be:
3, 3/10(x-3), 3/75(x-3)2, 3/189(x-3)3
So f(3) equals 3, I'm pretty sure.
Because all the other (except the 1st) cancel to equal 0.
3 + infinity zeros = 3
The second question asks me whether or not f(4) converges.
Well, that would be: 3 + (3/10)(1)1 + (3/75)(1)2+(3/189)(1)3 + ...
3 + increasingly small numbers = convergence
Is this the right way to think about it? Is this is correct, is there a better, perhaps faster way?
The third question asks whether or not f(8) converges.
That would be: 3 + (3/10)(5)1 + (3/75)(5)2 + (3/189)(5)3 + ...
= 3 + increasingly small numbers = convergence
...I don't think it's getting bigger very fast... It seems like it would take way too long for this to go to infinity. That's why I'm thinking it would converge...
The fourth question asks whether or not f(-2) converges.
That would be: 3 + (3/10)(5)1 + (3/75)(5)2 + ...
Yeah, this is the same as last time. So I say it converges again.
Now, the fifth question asks what the interval of convergence is for this power series.
To find this interval, I think I need to use the formula: an/an+1
[itex]\frac{3(x-3)^{n}}{(n+1)5^{n}}[/itex]*[itex]\frac{(n+2)(5^{n+1})}{3(x-3)^{n+1}}[/itex]
= [itex]\frac{5(n+2)}{(n+1)(x-3)}[/itex]
Now, the next step is to take the limit, but in this form, I'll get ∞/∞.
I could do L'Hospital's Rule, but I think it might be easier if I rewrote the problem before I take the limit.
Hopefully, I'm not doing any illegal math moves... :/
[itex]\frac{5}{(n+2)^{-1}(n+1)(x-3)}[/itex]
=[itex]\frac{5}{n^{0}+n^{-1}+2n^{-1}+2{-1}(x-3)}[/itex]
= [itex]\frac{5}{1 + 1/n + 1/2n + 1/2(x-3)}[/itex]
lim |[itex]\frac{5}{1.5 + 1/n + 1/2n(x-3)}[/itex]|
n→∞
= |5/[(1.5 + 0 + 0)(x-3)]|
I CAN'T FIND THE GREATER THAN SYMBOL, so I'm using "∠"!
= -1 ∠ 10/3(x-3) ∠ 1
= -33/10 ∠ x ∠ 33/10
...So it is convergent between this interval... Which I'm pretty sure is WRONG!
That is too hideous of an answer to be correct. AND- it does not fit with my convergence answers from before!
33/10= 3.3
4 AND 8 lie outside of this interval. Something's wrong here... :/
Please help! D=
Thanks :)