To find the interval of convergence for the series

$\sum_{n=1}^{\infty} \left( n^{\frac{1}{n}} - 1 \right) x^n,$

we can apply the Root Test because it is particularly effective when we have terms raised to the power $n$.

Step 1: Apply the Root Test

The Root Test states that for a series $\sum a_n x^n$, the radius of convergence $R$ can be determined using:

$\lim_{n \to \infty} \sqrt[n]{|a_n x^n|} = L.$

If $L < 1$, the series converges, and if $L > 1$, the series diverges. The radius of convergence $R$ is the reciprocal of this limit.

For our series, let $a_n = n^{\frac{1}{n}} - 1.$

Step 2: Evaluate $\sqrt[n]{|a_n x^n|}$

Notice that: [ \sqrt[n]{|a_n x^n|} = |x| \cdot \sqrt[n]{|a_n|}. ]

Therefore, we need to find: $\lim_{n \to \infty} |x| \cdot \sqrt[n]{|n^{\frac{1}{n}} - 1|}.$

Step 3: Evaluate $\lim_{n \to \infty} \sqrt[n]{|n^{\frac{1}{n}} - 1|}$

To determine the behavior of $n^{\frac{1}{n}} - 1$ as $n \to \infty$, observe that $n^{\frac{1}{n}} \to 1$ as $n \to \infty$. This implies that: $n^{\frac{1}{n}} - 1 \to 0 \text{ as } n \to \infty.$

Since $n^{\frac{1}{n}} \to 1$, the expression $n^{\frac{1}{n}} - 1$ approaches 0 extremely slowly. However, the key is that $\lim_{n \to \infty} \sqrt[n]{n^{\frac{1}{n}} - 1} = 1.$

Step 4: Apply the Root Test Result

Thus, $\lim_{n \to \infty} |x| \cdot \sqrt[n]{|a_n|} = |x| \cdot 1 = |x|.$

For convergence, we require $|x| < 1$. Therefore, the radius of convergence is $R = 1$, and the series converges for $|x| < 1$.

Step 5: Determine Convergence at the Endpoints

To find the interval of convergence, we must check the behavior of the series at $x = \pm 1$.

1. At $x = 1$:

   The series becomes $\sum_{n=1}^{\infty} \left( n^{\frac{1}{n}} - 1 \right).$

   Since $n^{\frac{1}{n}} - 1 \to 0$ as $n \to \infty$, the terms do not decrease quickly enough to ensure convergence. Thus, the series diverges at $x = 1$.

2. At $x = -1$:

   The series becomes $\sum_{n=1}^{\infty} \left( n^{\frac{1}{n}} - 1 \right) (-1)^n.$

   This is an alternating series, but again $n^{\frac{1}{n}} - 1$ does not tend to zero quickly enough to ensure convergence by the Alternating Series Test. Thus, the series also diverges at $x = -1$.

Conclusion

The interval of convergence is:

$(-1, 1).$

Would you like a deeper explanation on any of the steps?

Here are five additional questions that build on this concept:

1. How does the Root Test differ from the Ratio Test, and when is each most useful?
2. What happens if we change the general term $n^{\frac{1}{n}} - 1$ to a different function of $n$?
3. How can we determine the convergence of a power series if the Root and Ratio Tests both yield inconclusive results?
4. Why does the alternating series test not guarantee convergence at $x = -1$ in this case?
5. Can you derive the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{x^n}{n}$?

Tip: For series with complicated general terms, the Root Test is often simpler than the Ratio Test because it directly gives the radius of convergence.