Question

Consider the sequence \(\left \{ \frac{1}{k} \right \}_{k=0}^{\infty }\). The associated sequence \(\left \{ S_{n}\right \}_{n=0}^{\infty }\), 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 \(S_10\) 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 \(e\).