Question

Use the results of the ratio test.\\ $\sum_{n=1}^{\infty} \frac{16^n (x - 8)^n}{n!}$ converges absolutely for all x. Therefore, the radius of convergence is R = 0.\\ Also, the above series converges for all x in the interval.\\ Note that this implies there is/are 2 endpoint(s) to check.

          Use the results of the ratio test.\\
$\sum_{n=1}^{\infty} \frac{16^n (x - 8)^n}{n!}$ converges absolutely for all x. Therefore, the radius of convergence is R = 0.\\
Also, the above series converges for all x in the interval.\\
Note that this implies there is/are 2 endpoint(s) to check.

Use the results of the ratio test.

∑n=1^∞(16^n (x - 8)^n)/(n!) converges absolutely for all x. Therefore, the radius of convergence is R = 0.

Also, the above series converges for all x in the interval.

Note that this implies there is/are 2 endpoint(s) to check.

Added by Jose Carlos O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Andrew Noble

06/07/2023

Step 1

The ratio test states that for a series \sum_{n=1}^{\infty} a_n, if lim_{n \to \infty} |a_{n+1}/a_n| = L, then the series converges absolutely if L < 1 and diverges if L > 1. In this case, we have a_n = \frac{16^{n}(x-8)^{n}}{n!}. Let's calculate the ratio of Show more…

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Use the results of the ratio test: sum_{n=1}^{infty} frac{16^{n}(x-8)^{n}}{n!} converges absolutely for x. Therefore, the radius of convergence is R. Also, the above series converges for all x in the interval. Note that this implies there are endpoints to check.