b7-power series: problem 8\n(4 points)\nfind all the values of x such that the given series would…
b7-power series: problem 8\n(4 points)\nfind all the values of x such that the given series would converge.\n\\( \\sum _ { n = 1 } ^ { \\infty } \\frac { ( - 1 ) ^ { n } x ^ { n } } { 11 ^ { n } \\left( n ^ { 2 } + 2 \\right) } \\)\nthe series is convergent\nfrom \\( x = \\)\nleft end included (enter y or n):\nto \\( x = \\)\nright end included (enter y or n):\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 10 attempts remaining.\npage generated october 15, 2025, 4:26:21 pm edt\nwebwork \\( \\odot 1996 - 2025 | \\) theme: math4 | ww_version 2.20 | pg_version 2.20\nthe webwork project
Answer
Explanation:
Step1: Apply Ratio Test
For a power series (\sum_{n = 1}^{\infty}a_{n}), where (a_{n}=\frac{(- 1)^{n}x^{n}}{11^{n}(n^{2}+2)}), use the ratio test. Calculate (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|).
[ \begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right|&=\left|\frac{\frac{(-1)^{n + 1}x^{n+1}}{11^{n+1}((n + 1)^{2}+2)}}{\frac{(-1)^{n}x^{n}}{11^{n}(n^{2}+2)}}\right|\ &=\left|\frac{(-1)^{n+1}x^{n + 1}\times11^{n}(n^{2}+2)}{(-1)^{n}x^{n}\times11^{n+1}((n + 1)^{2}+2)}\right|\ &=\left|\frac{-x(n^{2}+2)}{11((n + 1)^{2}+2)}\right|\ &=\left|\frac{x}{11}\right|\lim_{n\rightarrow\infty}\frac{n^{2}+2}{n^{2}+2n + 3}\ \end{align*} ]
Since (\lim_{n\rightarrow\infty}\frac{n^{2}+2}{n^{2}+2n + 3}=1) (divide numerator and denominator by (n^{2}): (\lim_{n\rightarrow\infty}\frac{1+\frac{2}{n^{2}}}{1+\frac{2}{n}+\frac{3}{n^{2}}}=1)), the series converges when (\left|\frac{x}{11}\right|\lt1), i.e., (- 11\lt x\lt11).
Step2: Check the left - hand endpoint (x=-11)
When (x = - 11), the series becomes (\sum_{n=1}^{\infty}\frac{(-1)^{n}(-11)^{n}}{11^{n}(n^{2}+2)}=\sum_{n = 1}^{\infty}\frac{1}{n^{2}+2}).
Since (n^{2}+2\gt n^{2}) and (\sum_{n = 1}^{\infty}\frac{1}{n^{2}}) is a convergent (p -)series ((p = 2\gt1)), by the comparison test (\sum_{n=1}^{\infty}\frac{1}{n^{2}+2}) converges.
Step3: Check the right - hand endpoint (x = 11)
When (x = 11), the series becomes (\sum_{n=1}^{\infty}\frac{(-1)^{n}(11)^{n}}{11^{n}(n^{2}+2)}=\sum_{n = 1}^{\infty}\frac{(-1)^{n}}{n^{2}+2}).
This is an alternating series. Let (b_{n}=\frac{1}{n^{2}+2}). Then (b_{n+1}=\frac{1}{(n + 1)^{2}+2}\lt\frac{1}{n^{2}+2}=b_{n}) for all (n\geq1) and (\lim_{n\rightarrow\infty}b_{n}=\lim_{n\rightarrow\infty}\frac{1}{n^{2}+2}=0). By the Alternating Series Test, (\sum_{n = 1}^{\infty}\frac{(-1)^{n}}{n^{2}+2}) converges.
Answer:
from (x=-11), left end included ((Y)): to (x = 11), right end included ((Y)):