# 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!)

May 17, 2018

(a) $\frac{11}{4} \pi$
(b) $7 + {e}^{-} 8$

#### Explanation:

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

(a)

We know

$\ln \left(1 + z\right) = z - {z}^{2} / 2 + {z}^{3} / 3 - {z}^{4} / 4 + \ldots$

Hence

$\ln \left(1 - z\right) = - z - {z}^{2} / 2 - {z}^{3} / 3 - {z}^{4} / 4 + \ldots$

and so

$\ln \left(\frac{1 + z}{1 - z}\right) = 2 \left(z + {z}^{3} / 3 + {z}^{5} / 5 + \ldots\right)$

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

$\ln \left(\frac{1 + i}{1 - i}\right) = 2 i \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots\right)$

Thus

$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \frac{1}{2 i} \ln \left(\frac{1 + i}{1 - i}\right)$

Now

$\frac{1 + i}{1 - i} = {\left(1 + i\right)}^{2} / \left(1 - {i}^{2}\right) = \left(1 + 2 i + {i}^{2}\right) \left(1 - \left(- 1\right)\right) = i$

Thus

$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \frac{1}{2 i} \ln \left(i\right) = \frac{i \frac{\pi}{2}}{2 i} = \frac{\pi}{4}$

So the given series

$11 - \frac{11}{3} + \frac{11}{5} - \frac{11}{7} + \ldots = \frac{11}{4} \pi$

(b)

sum_(n=2)^oo ((-1)^n(8^n))/(n!) = sum_(n=0)^oo (-8)^n/(n!)-(-8)^0/(0!)-(-8)^1/(1!)
$q \quad = {e}^{-} 8 - 1 + 8 = 7 + {e}^{-} 8$