Question

How do you find the radius of convergence $ \sum {{n^n}{x^n}} $ from $ n = [1,\infty ) $ ?

Answer 1

Hint: In this question we are given an infinite series and we have to find the radius of its convergence. For that, we have to first understand what an infinite series is and what its convergence is. A series is defined as an expression in which infinitely many terms are added one after the other to a given starting quantity. It is represented as $ \sum\limits_{n = 1}^\infty {{a_n}} $ where $ \sum {} $ sign denotes the summation sign which indicates the addition of all the terms. When we get further and further in a sequence, the terms get closer and closer to a specific limit, this signifies the convergence of the series. To find the radius of convergence of a series, we have to first test whether the series actually converges or not.

Complete step by step solution:
The series given is $ \sum {{n^n}{x^n}} $ , so the nth term of this series is given as - $ {a_n} = {n^n}{x^n} $
Now we will do the ratio test of this series, we have $ L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| $
Putting the value of $ {a_n} $ in the above equation, we have –
\[
  L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{{(n + 1)}^{n + 1}}{x^{n + 1}}}}{{{n^n}{x^n}}}} \right| = L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{(n + 1){{(n + 1)}^n}{x^{n + 1}}}}{{{n^n}{x^n}}}} \right| \\
   \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } (n + 1){(\dfrac{{n + 1}}{n})^n}\left| x \right| \\
   \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } (n + 1){(1 + \dfrac{1}{n})^n}\left| x \right| \\
 \]
As n approaches infinity, $ n + 1 \to \infty $ and $ \dfrac{1}{n} $ approaches zero, so $ (1 + \dfrac{1}{n}) \to 1 $
We know that $ {(1)^\infty } = 1 $ , thus $ {(1 + \dfrac{1}{n})^n} \to 1 $ , $ \left| x \right| $ is a positive number for any value of x and infinity multiplied with any number is infinity, so –
\[
   \Rightarrow L = \infty (1)\left| x \right| \\
   \Rightarrow L = \infty \\
   \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \infty \\
 \]
Thus, the series doesn’t converge, instead it diverges.
Hence, the series $ \sum {{n^n}{x^n}} $ has zero radius of convergence.
So, the correct answer is “Hence, the series $ \sum {{n^n}{x^n}} $ has zero radius of convergence.

Note: To test the convergence of a series $ \sum\limits_{n = 1}^\infty {{a_n}} $ , we do a test called the ratio test, it is also known as D’Alembert’s ratio test or the Cauchy ratio test. Each term of the series is a real or a complex number, $ {a_n} $ is not equal to zero and n is a large value. The ratio test is given as $ L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| $ , if the value of L comes out to be smaller than 1 then the series converges, if it is equal to 1 then the test is inconclusive as the limit fails to exist and if the value comes out to be greater than 1 then the series is divergent.