Question

How do you determine the convergence or divergence of \[\sum{\dfrac{{{\left( -1 \right)}^{n}}}{n!}}\] from $\left[ 1,\infty \right)$ ?

Answer 1

Hint: To determine the convergence or divergence of \[\sum{\dfrac{{{\left( -1 \right)}^{n}}}{n!}}\] from $\left[ 1,\infty \right)$ , we will be using d'Alembert's ratio test. Let us consider a series $S=\sum\limits_{r=1}^{\infty }{{{a}_{n}}}$ and $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$ . If we obtain L < 1 then the series converges absolutely.If we get L > 1 then the series is divergent and if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case. We have to consider \[{{a}_{n}}=\dfrac{{{\left( -1 \right)}^{n}}}{n!}\] and find L.

Complete step by step solution:
We need to determine the convergence or divergence of \[\sum{\dfrac{{{\left( -1 \right)}^{n}}}{n!}}\] from $\left[ 1,\infty \right)$ . We will be using d'Alembert's ratio test. Let us consider a series $S=\sum\limits_{r=1}^{\infty }{{{a}_{n}}}$ and $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$ .
If we obtain L < 1 then the series converges absolutely.If we get L > 1 then the series is divergent and if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.
Now, let us consider \[{{a}_{n}}=\dfrac{{{\left( -1 \right)}^{n}}}{n!}\] . Let’s find L.
$\begin{align}
  & L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( -1 \right)}^{n+1}}}{\left( n+1 \right)!}}{\dfrac{{{\left( -1 \right)}^{n}}}{n!}} \right| \\
 & \Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{\left( -1 \right)}^{n+1}}}{\left( n+1 \right)!}\times \dfrac{n!}{{{\left( -1 \right)}^{n}}} \right| \\
\end{align}$
Let us expand $\left( n+1 \right)!$ as follows.
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{\left( -1 \right)}^{n+1}}}{\left( n+1 \right)\times n!}\times \dfrac{n!}{{{\left( -1 \right)}^{n}}} \right|\]
Now, we can cancel the common terms.
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{\left( -1 \right)}^{n+1}}}{\left( n+1 \right)}\times \dfrac{1}{{{\left( -1 \right)}^{n}}} \right|\]
We know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ . Hence, the above equation becomes
\[\begin{align}
  & \Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{1}{\left( n+1 \right)}\times \dfrac{{{\left( -1 \right)}^{n+1}}}{{{\left( -1 \right)}^{n}}} \right| \\
 & \Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{1}{\left( n+1 \right)}\times -1 \right| \\
 & \Rightarrow L=\displaystyle \lim_{n \to \infty }\left| -\dfrac{1}{\left( n+1 \right)} \right| \\
\end{align}\]
We can write the above equation as
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\dfrac{1}{\left( n+1 \right)}\]
Let us now apply the limit. We will get
\[L=\dfrac{1}{\left( \infty +1 \right)}=\dfrac{1}{\infty }=0\]
We can see that L<1.
Therefore, the given series converges absolutely.

Note: Students must know the rules of exponents and factorials to solve these types of problems. Students may make mistakes by writing the formula for L as $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n}}}{{{a}_{n+1}}} \right|$ . Before applying the limits, we must take the absolute value of the expression as in \[L=\displaystyle \lim_{n \to \infty }\left| -\dfrac{1}{\left( n+1 \right)} \right|\] .