Question

A series of monomials: Use the ratio test for absolute convergence to find all values of x for which the series$\sum _{k=1}^{\infty }\frac{x^k}{k!}$converges.

Short answer

The value of x is less than k

Step-by-step solution

Step 1. Given 

The given series is$\sum _{k=1}^{\infty }\frac{x^k}{k!}$

Step 2. Ratio test 

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{RatioTest},\underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{a}}_{\mathrm{k}}}}\mathrm{be}\mathrm{the}\mathrm{series}\mathrm{with}\mathrm{positive}\mathrm{terms},\mathrm{if}\mathrm{L}=\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}+1}}{{\mathrm{a}}_{\mathrm{k}}} \\ \mathrm{If}\mathrm{aL}<1\mathrm{series}\mathrm{converges}. \\ 2.\mathrm{If}\mathrm{L}>1\mathrm{series}\mathrm{diverges}. \\ 3.\mathrm{If}\mathrm{L}=1\mathrm{the}\mathrm{test}\mathrm{is}\mathrm{inconclusive}. \end{gathered}$$

Step 3. Finding the value of x.

$$\begin{gathered} \mathrm{Now},\mathrm{calculate}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{Now},\mathrm{calculate}\mathrm{the}\mathrm{value}\mathrm{of}\frac{a_{k+1}}{a_k} \\ \mathrm{Consider}\mathrm{the}\mathrm{general}\mathrm{term}as=\frac{x^k}{\left(k\right)!}. \\ \mathrm{Therefore},\mathrm{by}\mathrm{substituting}\mathrm{k}=\mathrm{k}+1 \\ {a}_{k+1}=\frac{x^{(k+1)}}{(k+1)!} \\ Now,\frac{a_{k+1}}{a_k}=\frac{{\displaystyle \frac{x^{k+1}}{(k+1)!}}}{x^k/k!} \\ \mathrm{Using}\mathrm{formula}\mathrm{n}!=n(n-1)! \\ \frac{a_{k+1}}{a_k}=\frac{x}{k} \\ Takinglimits \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\left(\frac{x}{k}\right) \\ x<k. \\ \mathrm{for}\mathrm{all}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{less}\mathrm{than}\mathrm{k},\mathrm{the}\mathrm{series}\mathrm{is}\mathrm{convergent} \end{gathered}$$