Question

explain why it would be difficult to use the root test on the series$\sum _{k=1}^{\infty }\frac{1}{k!}$

Short answer

The root test is difficult and ratio test is easy to do the convergence test

Step-by-step solution

Given information

Consider the given series,

$\sum _{k=1}^{\infty }\frac{1}{k!}$

Calculation.

Consider the series $\sum _{k=1}^{\infty }\frac{1}{k!}$

The purpose is to clarify why applying the Root test to the provided series is challenging.

Follow the steps as outlined to determine why using the Root Test is challenging.

Now, according to Root Test $\sum _{k=1}^{\infty }{a}_k$be the series with all terms positive and $L=\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\frac{1}{i}}$then,

1. If $L<1$series converges.

2. If $L>1$series diverges.

3. If $L=1$the test is inconclusive.

The general term of the series is $a_k=\frac{1}{k!}$.

Calculate the value $L=\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\frac{1}{4}}$.

$\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\frac{1}{k}}=\underset{k\to \infty }{\lim }{\left(\frac{1}{k!}\right)}^{\frac{1}{k}}\dots \dots \left(1\right)$

Now, it is challenging to compute the limit in (1). Consequently, it is challenging to use the Root test.

Further simplification

Now, according to Root Test $\sum _{k=1}^{\infty }{a}_k$be the series with all terms positive and $L=\underset{k\to \infty }{\lim }\left(\frac{a_{k+1}}{a_k}\right)$then,

1. If $L<1$series converges.

2. If $L>1$series diverges.

3. If $L=1$the test is inconclusive.

To find the value of $L=\underset{k\to \infty }{\lim }\left(\frac{a_{k+1}}{a_k}\right)$

localid="1661335075977" $\begin{array}{rcl}\underset{k\to \infty }{\lim }\left(\frac{a_{k+1}}{a_k}\right)& =& \underset{k\to \infty }{\lim }\frac{\frac{1}{(k+1)!}}{\frac{1}{k!}}\\ & =& \underset{k\to \infty }{\lim }\frac{k!}{(k+1)!}\end{array}$

Use $n!=n(n-1)!$and simplify.

localid="1661335084042" $\begin{array}{rcl}\underset{k\to \infty }{\lim }\left(\frac{a_{k+1}}{a_k}\right)& =& \underset{k\to \infty }{\lim }\frac{\backslash \mathrm{not}k}{(k+1)\backslash \mathrm{not}!}\\ & =& \underset{k\to \infty }{\lim }\frac{1}{(k+1)}\\ & =& 0\end{array}$

Thus,$L=0$

Now, $L<1$thus, the series is convergent.

The root test is difficult and ratio test is easy to do as the convergence test.