Find the radius of convergence, R, of the series.

Question

∑ (from n=1 to ∞) (n^2 x^n)/(7 · 14 · 21 ·    · (7n))

R = ?

Find the interval, i, of convergence of the series. (Enter your answer using interval notation.)

i = ?

Answer 1

To find the radius of convergence of the series

∑n=1∞n2xn7⋅14⋅21⋯(7n),\sum_{n=1}^{\infty} \frac{n^2 x^n}{7 \cdot 14 \cdot 21 \cdots (7n)},n=1∑∞​7⋅14⋅21⋯(7n)n2xn​,

we start by expressing the general term of the series:

an=n2xn7⋅14⋅21⋯(7n).a_n = \frac{n^2 x^n}{7 \cdot 14 \cdot 21 \cdots (7n)}.an​=7⋅14⋅21⋯(7n)n2xn​.

The denominator can be written as a product:

7⋅14⋅21⋯(7n)=7n⋅(1⋅2⋅3⋯n)=7n⋅n!,7 \cdot 14 \cdot 21 \cdots (7n) = 7^n \cdot (1 \cdot 2 \cdot 3 \cdots n) = 7^n \cdot n!,7⋅14⋅21⋯(7n)=7n⋅(1⋅2⋅3⋯n)=7n⋅n!,

thus,

a_n=n2xn7n⋅n!.a_n = \frac{n^2 x^n}{7^n \cdot n!}.an​=7n⋅n!n2xn​.

Now we apply the ratio test, which involves finding the limit of the absolute value of the ratio of successive terms:

L=lim⁡n→∞∣an+1an∣.L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.L=n→∞lim​​an​an+1​​​.

Let's compute this ratio:

an+1=(n+1)2xn+17n+1⋅(n+1)!.a_{n+1} = \frac{(n+1)^2 x^{n+1}}{7^{n+1} \cdot (n+1)!}.an+1​=7n+1⋅(n+1)!(n+1)2xn+1​.

So,

∣an+1an∣=∣(n+1)2xn+17n+1(n+1)!⋅7n⋅n!n2xn∣.\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)^2 x^{n+1}}{7^{n+1} (n+1)!} \cdot \frac{7^n \cdot n!}{n^2 x^n} \right|.​an​an+1​​​=​7n+1(n+1)!(n+1)2xn+1​⋅n2xn7n⋅n!​​.

Simplifying this, we get:

∣an+1an∣=∣(n+1)2xn+17n+1(n+1)!⋅7n⋅n!n2xn∣=∣(n+1)2x7(n+1)n2∣.\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)^2 x^{n+1}}{7^{n+1} (n+1)!} \cdot \frac{7^n \cdot n!}{n^2 x^n} \right| = \left| \frac{(n+1)^2 x}{7 (n+1) n^2} \right|.​an​an+1​​​=​7n+1(n+1)!(n+1)2xn+1​⋅n2xn7n⋅n!​​=​7(n+1)n2(n+1)2x​​.

As n→∞n \to \inftyn→∞,

∣n+17n2x∣≈∣x7n∣.\left| \frac{n+1}{7 n^2} x \right| \approx \left| \frac{x}{7n} \right|.​7n2n+1​x​≈​7nx​​.

Taking the limit as n→∞n \to \inftyn→∞,

L=lim⁡n→∞∣x7n∣=0.L = \lim_{n \to \infty} \left| \frac{x}{7n} \right| = 0.L=n→∞lim​​7nx​​=0.

Since L=0for all x, the series converges for all x. Therefore, the radius of convergence R is:

R=∞.