Why Does the Series Sum of n1 to Infinity of 1/n! Converge?

In this article, we will delve into the convergence of the series (sum_{n1}^{infty} frac{1}{n!}) and explain why it converges using various analytical methods. This series is quite intriguing due to its relationship with the mathematical constant (e).

Introduction to the Series

The series in question is (sum_{n1}^{infty} frac{1}{n!}). This series represents a sum of the reciprocals of factorials of natural numbers starting from 1. It is a fascinating series that appears in many areas of mathematics, particularly in calculus and analysis.

Convergence with the Ratio Test

One of the most powerful tools to determine the convergence of a series is the (text{ratio test}). The ratio test states that for a series (sum a_n), if we let (L lim_{n to infty} left| frac{a_{n 1}}{a_n} right|) exist, then:

If (L 1), the series converges absolutely. If (L 1), the series diverges. If (L 1), the test is inconclusive.

Applying the Ratio Test to the Series

For our series, let (a_n frac{1}{n!}). We need to find the ratio (frac{a_{n 1}}{a_n}):

(a_{n 1} frac{1}{(n 1)!} frac{1}{(n 1) cdot n!}) (frac{a_{n 1}}{a_n} frac{frac{1}{(n 1)!}}{frac{1}{n!}} frac{1}{(n 1) cdot n!} cdot n! frac{1}{n 1}) (L lim_{n to infty} left| frac{1}{n 1} right|) (L 0 1)

Since (L 0 1), the series converges absolutely by the ratio test.

Alternative Method: Taylor Series Expansion

We can also determine the convergence of the series (sum_{n1}^{infty} frac{1}{n!}) by recognizing it as a part of the Taylor series expansion of the function (e^x) around (x 0) and evaluating it at (x 1).

Taylor Series Expansion

The Taylor series expansion for the exponential function is given by:

(e^x sum_{n0}^{infty} frac{x^n}{n!})

Substituting (x 1) into the series, we get:

(e^1 sum_{n0}^{infty} frac{1^n}{n!} sum_{n0}^{infty} frac{1}{n!})

The series (sum_{n0}^{infty} frac{1}{n!}) includes a term for (n 0), which is 1. Therefore, the series (sum_{n1}^{infty} frac{1}{n!}) is equal to:

(e^1 - 1 e - 1)

Hence, the series converges to (e - 1).

Comparison and Dominance

To further understand the convergence, we can use the comparison test. Notice that the series (sum_{n1}^{infty} frac{1}{n!}) is dominated by the series (sum_{n1}^{infty} frac{1}{n^2}), which is a well-known convergent series. Since the larger series converges, so does the smaller one by the comparison test.

The series (sum_{n1}^{infty} frac{1}{n^2}) is a p-series with (p 2 1), which converges to (frac{pi^2}{6}). Thus, (sum_{n1}^{infty} frac{1}{n!}) also converges as its terms are eventually smaller than those of the p-series.

Conclusion

In conclusion, the series (sum_{n1}^{infty} frac{1}{n!}) converges not only because it satisfies the ratio test but also through its relation to the Taylor series expansion of (e^x). Ultimately, the series converges to (e - 1), highlighting the deep connections between different mathematical concepts.

Thank you for reading, and if you have any questions or further insights, feel free to comment below!