How do you find the radius of convergence #Sigma (n^nx^n)/(lnn)^n# from #n=[2,oo)#?

Question

Ezekiel Alexander · Accepted Answer

The root test states that the series #suma_n# is convergent if #L=lim_(nrarroo)root(n)abs(a_n)<1#. Here we see that #a_n=((nx)/lnn)^n#, so: #L=lim_(nrarroo)root(n)abs(((nx)/lnn)^n)=lim_(nrarroo)abs((xn)/lnn)# Since the limit is dependent only on the change in #n#, the #x# can be removed from the limit like a constant: #L=absxlim_(nrarroo)abs(n/lnn)# We see that the limit approaches #oo#, since #n# grows faster than the logarithmic function #lnn#. The only time when #L<1#, when the series converges, will be when #L=0#, which occurs only when #x=0#. Since there is only one value of #x# for which the series converges, we see that #R=0#.

Aurora Alvarado · Answer

undefined To find the radius of convergence of the series $\sum_{n=2}^{\infty} \frac{n^n x^n}{(\ln n)^n}$, you can apply the ratio test.

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

Here, $a_n = \frac{n^n x^n}{(\ln n)^n}$.

Now, compute $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$:

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

Simplify and apply limit properties:

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

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

$$
= \lim_{n \to \infty} \left| \frac{(n+1)(\ln n)}{(\ln (n+1))} \right|
$$

Apply L'Hôpital's rule:

$$
= \lim_{n \to \infty} \left| \frac{\frac{d}{dn}(n+1)(\ln n)}{\frac{d}{dn}(\ln (n+1))} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{\frac{1}{n+1} \cdot (\ln n) + (n+1) \cdot \frac{1}{n}}{\frac{1}{n+1}} \right|
$$

$$
= \lim_{n \to \infty} \left| (\ln n) + \frac{n+1}{n} \right|
$$

$$
= \lim_{n \to \infty} \left| \ln n + 1 + \frac{1}{n} \right|
$$

As $n$ tends to infinity, $\frac{1}{n}$ approaches 0. So, the limit is:

$$
= \lim_{n \to \infty} \left| \ln n + 1 \right| = \infty
$$

Since the limit is infinity, the series diverges for all $x \neq 0$. Therefore, the radius of convergence is $R = 0$.