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!) #

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

(a)

We know

Hence

and so

Thus

Now

Thus

So the given series

(b)

(a) To evaluate the series ( \frac{11}{1} - \frac{11}{3} + \frac{11}{5} - \frac{11}{7} + \frac{11}{9} - \frac{11}{11} + \ldots ), we observe that it is an alternating series. By applying the Alternating Series Test, we can determine its convergence. The absolute value of each term in the series decreases as the index increases, and the limit of the absolute value of the terms as ( n ) approaches infinity is zero. Therefore, the series converges. To find its sum, we can express it as a power series. The series can be rewritten as ( \sum_{n=0}^{\infty} (-1)^n \frac{11}{2n+1} ). This is a power series representation of the function ( f(x) = 11 \arctan(x) ). So, the sum of the series is ( 11 \arctan(1) = \frac{11\pi}{4} ).

(b) The series ( \sum_{n=2}^{\infty} \frac{(-1)^n \cdot 8^n}{n!} ) can be recognized as an alternating series by noting the ( (-1)^n ) term. We can apply the Ratio Test to determine its convergence. The absolute value of the ratio of successive terms is ( \frac{8}{n+1} ), which approaches zero as ( n ) approaches infinity. Therefore, the series converges. To find its sum, we recognize it as the Maclaurin series expansion of ( e^{-8} ). Thus, the sum of the series is ( e^{-8} ).

(a) The given series is a telescoping series. By examining the pattern, we see that it alternates between positive and negative terms and that each term cancels out all but one term from the previous and next pairs. So, the sum of the series is the value of the first term which is 11/1.

(b) This is the Maclaurin series expansion of the function ( e^{-8} ). Therefore, the sum of the series is ( e^{-8} ).