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.

1 Answer
May 17, 2018

(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#