General Forums [Trash Bin]Come on peeps get with the name. :P It's basically a place to move all the dead topics over here... However, To keep people's posts around. The topics here are not deleted. o.o
To find the interval of convergence I started with the ratio test. (limit as n->∞ for the absolute value of (an+1/an). If the result is less than 1 the series converges, if it is greater than 1 it diverges. (Test is inconclusive if it is equal to 1)
So proceeding with the test, the limit as n->∞ of
| (x-2)^(n+1) * n(-3)^n____| // I just multiplied by the reciprocal of an
| (n+1)(-3)^(n+1) * (x-2)^n |
You then cancel out everything you can and get…
the limit as n->∞ of
|(x-2) * n |
|-3 *(n+1)|
The limit of n/(n+1) as n-> ∞ is 1 so it can be ignored.
So the series converges when …
|(x-2) / -3| < 1
This is broken up into two parts…
(x-2)/-3 <1
x-2 > -3
x > -1
(x-2)/-3 > -1
x < 5
now we know that the series converges between -1 and 5 (eliminating the popular answer choice E). now we need to check the endpoints.
Substitute in the value for of the endpoints in for the x in the original
∞
Σ ( (-1-2)^n ) / ( n * (-3)^n)
n=1
This simplifies down to
∞
Σ ( (-3)^n / ( n * (-3)^n)
n=1
Which equals
∞
Σ 1/n
n=1
Which diverges because of the p series test where
∞
Σ 1/n^p
n=1
If 0<p<1 the series diverges
So
∞
Σ 1/n
n=1
Diverges and our series thus diverges at x=-1
∞
Σ ( (5-2)^n ) / ( n * (-3)^n)
n=1
Equals…
∞
Σ ((-1)^n*3^n) / (n *3^n)
n=1
This simplifies to …
∞
Σ ((-1)^n) / (n)
n=1
Which converges due to the alternating series test,
Because its limit as n goes to infinity is 0 and its absolute value decreases as it goes to infinity
This is only true if the series alternates between positive and negative like this one.
So x=5 converges
Therefore the correct answer is C -1<x<5
Congratulations EinarRhys, the only one to get this one correct.
edit: after further review theeone also gets credit for answering correctly