How do you find the radius of convergence #Sigma (1*4*7* * * (3n+1))/(n!)x^n# from #n=[0,oo)#?

Question

Charles Arocha · Accepted Answer

The radius of convergence is #R= 1/3#

The ratio test states that a necessary condition for #sum_(n=0)^oo a_n# to converge is that:

#L = lim_(n->oo) abs(a_(n+1)/a_n) <= 1#

if #L<1# the condition is also sufficient and the series converges absolutely, while for #L=1# the test is inconclusive.

We then evaluate the ratio for the series at hand:

#abs(a_(n+1)/a_n) = (abs(x)^(n+1)(prod_1^(n+1) (3k+1))/((n+1)!))/(abs(x)^n(prod_1^n (3k+1))/(n!))= abs(x)(3(n+1) +1)/(n+1) = abs(x) (3n+4)/(n+1)#

so that:

#lim_(n->oo) abs(a_(n+1)/a_n) = 3abs(x)#

The series is then absolutely convergent for #absx <1/3# and divergent for #absx >1/3#with means that the radius of convergence is #R= 1/3#

Emilia Aguilar · Answer

undefined The radius of convergence of the series $ \sum_{n=0}^\infty \frac{(1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n+1))}{n!}x^n $ can be found using the ratio test.

The ratio test states that if $ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $, then the series converges absolutely if $ L < 1 $ and diverges if $ L > 1 $. If $ L = 1 $, the test is inconclusive.

In this case, $ a_n = \frac{(1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n+1))}{n!}x^n $.

Taking the ratio of consecutive terms:

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n+1))}{(n+1)!}x^{n+1}}{\frac{(1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n+1))}{n!}x^n} $$

$$ = \frac{(3n+4)x}{(n+1)(3n+1)} $$

Taking the limit as $ n $ approaches infinity:

$$ L = \lim_{n \to \infty} \left| \frac{(3n+4)x}{(n+1)(3n+1)} \right| $$

$$ = |3x| $$

For convergence, $ |3x| < 1 $, so $ |x| < \frac{1}{3} $. Therefore, the radius of convergence is $ \frac{1}{3} $.