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) $sum_(n=2)^oo ((-1)^n(8^n))/(n!)$

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

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Answer 1 · 2018-05-17T01:12:27

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

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)

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