Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(1+\frac{k}{n}\right)^{n} $$

Short answer

The series converges if \( e^k \) is smaller than 1 (which means k is a negative integer), diverges if \( e^k \) is greater than 1 (which means k is a positive integer), and the test is inconclusive if \( e^k \) is equal to 1 (which means k is 0).

Step-by-step solution

Recognize Euler's Number

First, recognize that \( \left(1+\frac{k}{n}\right)^{n} \) can be assumed to resemble the definition of Euler's number \( e^{k} \) in the limit as n approaches infinity. Recall that the number 'e' can be defined as \( e = \lim_{{n \to \infty}} \left( 1 + \frac{1}{n}\right)^n \), so for \( k \neq 0 \), \( e^{k} = \lim_{{n \to \infty}} \left( 1 + \frac{k}{n}\right)^n \)

Use the Root Test

Next, apply the Root Test, which is useful to determine the convergence of a series whose terms can be expressed as something to the power of 'n'. The Root Test says that a series \( \sum a_n \) converges if \( \lim_{{n \to \infty}} \sqrt[n]{|a_n|} < 1 \), diverges if \( \lim_{{n \to \infty}} \sqrt[n]{|a_n|} > 1 \), and is inconclusive if \( \lim_{{n \to \infty}} \sqrt[n]{|a_n|} = 1 \). Let's see if it can be applied to the given series.

Apply Root Test to the Series

Applying the Root Test, each term in the series becomes the absolute value of \( \lim_{{n \to \infty}} \sqrt[n]{\left( 1 + \frac{k}{n}\right)^n} \), which is simply the absolute value of \( e^k \). Since the value of 'k' doesn't change, the limit is continuously \( e^k \).

Assess the Result of the Test

Assess the result of the root test: if \( e^k \) is smaller than 1, then the series converges, if \( e^k \) is greater than 1, then the series diverges, and if \( e^k \) is equal to 1 (which means k is 0), the test is inconclusive.