Question

Consider the sequence ${\left\{\frac{1}{k}\right\}}_{k=0}^{\mathrm{\infty }}$. The associated sequence ${\left\{S_n\right\}}_{n=0}^{\mathrm{\infty }}$, where

$S_n=1+1+\frac{1}{2!}+...+\frac{1}{n!}$

, is a sequence of sums. In Chapter $8$ we will see that this sequence converges to the number $e$. Evaluate $S_n$ for $n=1,2,3,10$. How close is $S_{1}0$ to $e$?

Short answer

For the sequence $S_n=1+1+\frac{1}{2!}+...+\frac{1}{n!}$:

$S_1=2$,

$S_2=2.5$,

$S_3=2.667$,

$S_{10}=2.71858$.

The value of $S_{10}$ is approximately $0.0003$ more than the $e$.

Step-by-step solution

Step 1. Given Information

It is given that $S_n=1+1+\frac{1}{2!}+...+\frac{1}{n!}$, is a sequence of sums.

The objective is to evaluate $S_n$ for $n=1,2,3,10$ and to know how close is $S_{10}$ to $e$.

Step 2. Find $S_1$

For $n=1$ the sequence $S_n=1+1+\frac{1}{2!}+...+\frac{1}{n!}$ can be evaluated as:

$\begin{array}{rcl}{S}_1& =& 1+\frac{1}{1!}\\ & =& 1+1\\ & =& 2\end{array}$

Step 3. Find $S_2$

For $n=2$ the sequence $S_n=1+1+\frac{1}{2!}+...+\frac{1}{n!}$ can be evaluated as:

$\begin{array}{rcl}{S}_2& =& 1+1+\frac{1}{2!}\\ & =& 1+1+0.5\\ & =& 2.5\end{array}$

Step 4. Find $S_3$

For $n=3$ the sequence $S_n=1+1+\frac{1}{2!}+...+\frac{1}{n!}$ can be evaluated as:

$\begin{array}{rcl}{S}_3& =& 1+1+\frac{1}{2!}+\frac{1}{3!}\\ & =& 1+1+0.5+\frac{1}{6}\\ & \approx & 2.5+0.167\\ & =& 2.667\end{array}$

Step 5. Find $S_{10}$

For $n=10$ the sequence $S_n=1+1+\frac{1}{2!}+...+\frac{1}{n!}$ can be evaluated as:

$\begin{array}{rcl}{S}_{10}& =& 1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\frac{1}{9!}+\frac{1}{10!}\\ & =& 2+0.5+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}+\frac{1}{5040}+\frac{1}{40320}+\frac{1}{362880}+\frac{1}{3628800}\\ & \approx & 2.5+0.167+0.04167+0.0083+0.001389+0.000198+0.000025+0.00000276+0.000000276\\ & =& 2.71858\end{array}$

Step 6. How close is $S_{10}$ to $e$.

The value of $S_{10}$ is $2.71858$ and the value of $e$ is approximately $2.17828$.

$\begin{array}{rcl}{S}_{10}-e& =& 2.71858-2.71828\\ & =& 0.0003\end{array}$

So, $S_{10}$ is very close to $e$.