What is the interval of Convergence?

Question

For Σ (2^n(x+4)^n) / n (for n=0 --> ∞)

I got -9/2 and and -7/2, but is it (-9/2, -7/2) or [-9/2, -7/2)?

nle · Accepted Answer

You might miss the absolute value.

It will converge if  -9/2 < x < -7/2

Now you have to consider x = -9/2 and x = -7/2 separately.

If x = -9/2 it still converges but n must be from 1  to infinity.

If x = -7/2 it diverges, I ' ll prove it if I can


EDIT: OK, I'm back

If x = -9/2 then you can write:

a =  (2^n) * ( 4 - (  9/ 2) ) ^n / (n) 

or a= (2^n) * ( -1/2 ) ^n /n

a= (2^n) * ( (-1)^n  / 2^n) /n

which simplifies to a = (-1)^n /n

then the sum is  ( -1 + 1/2  - 1/3  + ..... )

which can be written:

S=  - ( 1 -1/2 + 1/3 - 1/4  + .. ....)

S is the negative of the famous alternating series

S1 = 1 -1/2 +1/ 3  - 1/4  + .....)

We know that S1 converges to ln(2)  (review your book)

then S converges to   - ln(2).

Therefore, the given series converges for

-  9/2 < =  x <  -7/2

Anonymous · Answer

just spin the top.

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What is the interval of Convergence?

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