Question

Calc 2 questions super lost! Evaluate a power series to find the sum of the series, or show that the series diverges. (If a series diverges, enter DIVERGES.)? (a) `11/1-11/3+11/5-11/7+11/9-11/11+....` (b)`sum_(n=2)^oo ((-1)^n(8^n))/(n!)`

please help not sure how to do it. I know they are both convergent.

Answer 1

(a) `11/4 pi`
(b) `7+e^-8`

Explanation:

Both of these series converge, as can be easily seen from the standard tests.

(a)

We know

`ln(1+z) = z-z^2/2+z^3/3-z^4/4+...`

Hence

`ln(1-z) = -z-z^2/2-z^3/3-z^4/4+...`

and so

`ln((1+z)/(1-z)) = 2(z+z^3/3+z^5/5+...)`

Substituting `z=i` in the above , we get

`ln((1+i)/(1-i)) = 2i(1-1/3+1/5-1/7+...)`

Thus

`1-1/3+1/5-1/7+... = 1/(2i)ln((1+i)/(1-i))`

Now

`(1+i)/(1-i) = (1+i)^2/(1-i^2)=(1+2i+i^2)(1-(-1)) = i`

Thus

`1-1/3+1/5-1/7+... = 1/(2i)ln(i) = (ipi/2)/(2i) = pi/4`

So the given series

`11-11/3+11/5-11/7+... =11/4 pi`

(b)

`sum_(n=2)^oo ((-1)^n(8^n))/(n!) = sum_(n=0)^oo (-8)^n/(n!)-(-8)^0/(0!)-(-8)^1/(1!)`
`qquad = e^-8-1+8=7+e^-8`