How do you find the radius of convergence #Sigma (x^n)/(5^(nsqrtn))# from #n=[0,oo)#?

Question

Emma Aldridge · Accepted Answer

The radius of convergence for the series:

#sum_(n=0)^oo x^n/(5^(nsqrtn))#

is #R=5#.

Evaluate the ratio: #abs( a_(n+1)/a_n) = abs ( ( x^(n+1)/(5^((n+1)sqrt(n+1))))/(x^n/(5^(nsqrtn))))# #abs( a_(n+1)/a_n) = abs ( x^(n+1)/x^n)5^(nsqrtn)/(5^((n+1)sqrt(n+1))# #abs( a_(n+1)/a_n) = abs ( x)5^n/5^(n+1)(5^sqrtn)/(5^(sqrt(n+1))# #abs( a_(n+1)/a_n) = abs ( x)/5 5^(sqrt(n)-sqrt(n+1))# Now note that: #lim_(n->oo) (sqrt(n)-sqrt(n+1)) = lim_(n->oo) (sqrt(n)-sqrt(n+1)) xx (sqrt(n)+sqrt(n+1))/ (sqrt(n)+sqrt(n+1))# #lim_(n->oo) (sqrt(n)-sqrt(n+1)) = lim_(n->oo) (n-(n+1))/ (sqrt(n)+sqrt(n+1))# #lim_(n->oo) (sqrt(n)-sqrt(n+1)) = lim_(n->oo) -1/ (sqrt(n)+sqrt(n+1)) = 0# so that: #lim_(n->oo) abs( a_(n+1)/a_n) = abs ( x)/5 lim_(n->oo)5^(sqrt(n)-sqrt(n+1)) = abs ( x)/5 xx 5^0 = abs ( x)/5# Based on the ratio test the series is therefore absolutely convergent for: #abs ( x)/5 < 1# and not convergent for #abs ( x)/5 > 1# So the radius of convergence is #R=5#

Evelyn Agard · Answer

undefined To find the radius of convergence of the series $ \sum_{n=0}^{\infty} \frac{x^n}{5^{n\sqrt{n}}} $, we use the ratio test.

Applying the ratio test, we have:

$$
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1} \cdot 5^{n\sqrt{n}}}{5^{(n+1)\sqrt{n+1}} \cdot x^n} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x \cdot 5^{n\sqrt{n}}}{5^{(n+1)\sqrt{n+1}}} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x}{5^{n\sqrt{n+1} - n\sqrt{n}}} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x}{5^{n(\sqrt{n+1} - \sqrt{n})}} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x}{5^{n\left(\frac{\sqrt{n+1} - \sqrt{n}}{1}\right)}} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x}{5^{n(\frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}})}} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x}{5^{n(\frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}})}} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x}{5^{n(\frac{1}{\sqrt{n+1} + \sqrt{n}})}} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{x}{5^{n(\frac{1}{\sqrt{n}(\sqrt{1+\frac{1}{n}} + 1)}})} \right|
$$

As $ n \to \infty $, $ \frac{1}{\sqrt{n}(\sqrt{1+\frac{1}{n}} + 1)} $ approaches $ \frac{1}{2} $. Therefore,

$$
\lim_{n \to \infty} \left| \frac{x}{5^{n(\frac{1}{\sqrt{n}(\sqrt{1+\frac{1}{n}} + 1)}})} \right| = \left| \frac{x}{5^{\frac{n}{2}}} \right|
$$

The series converges if this limit is less than 1, so we need:

$$
\left| \frac{x}{5^{\frac{n}{2}}} \right| < 1
$$

which simplifies to:

$$
|x| < 5^{\frac{n}{2}}
$$

Since $ 5^{\frac{n}{2}} $ increases as $ n $ increases, it converges for all $ x $ such that $ |x| < \lim_{n \to \infty} 5^{\frac{n}{2}} = \infty $, which means the radius of convergence $ R = \infty $.