The sum of the following infinite series is:
$ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... $

Question

The sum of the following infinite series is:$ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... $

Prepp Team · Accepted Answer

Infinite Series Sum Calculation

The question asks for the sum of the infinite series:

$ S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... $

Recall the Taylor series expansion for the exponential function '$e^x$' around $x=0$:

$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... $

Let $x=1$. The series becomes:

$ e^1 = e = \sum_{n=0}^{\infty} \frac{1^n}{n!} $

Since $1^n = 1$ for all $n$ and $0! = 1$, we have:

$ e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... $

$ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... $

The given series $ S $ is:

$ S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ... $

Comparing the expansion of '$e$' with the series '$S$', we see:

$ e = 1 + S $

Rearranging the equation to solve for $S$:

$ S = e - 1 $

Therefore, the sum of the infinite series is $ e - 1 $.

Prepp Team · Answer

$ \pi $

Prepp Team · Answer

$ 1 + e $

Prepp Team · Answer

$ e $

List-1	List-2
P The sum of the series $\sum_{n=1}^\infty \frac{1}{(n+2)(n+1)}$ is equal to	I $\frac{3}{2}$
Q $\lim_{x \to 0} \left( \frac{3}{x^2} \int_0^x \sin(t) dt \right)$ is equal to	II $1$
R Let $\frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$ be the Fourier series expansion of the function $f(x) = \frac{1}{2} \sin x - \frac{1}{2} \cos x + \frac{1}{\sqrt{2}} \sin 2x, x \in [0, 2\pi]$. Then, $\sum_{n=0}^\infty (a_n^2 + b_n^2)$ is equal to	III $\frac{1}{2}$