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Artist 52 [7]
3 years ago
12

Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

Mathematics
1 answer:
defon3 years ago
3 0
I'm guessing the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}

By the ratio test, the series converges if the following limit is less than 1.

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|

The first n terms in the numerator's denominator cancel with the denominator's denominator:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|

|x^n| also cancels out and the remaining factor of |x| can be pulled out of the limit (as it doesn't depend on n).

\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0

which means the series converges everywhere (independently of x), and so the radius of convergence is infinite.
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