First, we find the value of r:
[tex]\begin{gathered} a_n=\frac{2^n}{5^{n+1}\cdot n} \\ a_{n+1}=\frac{2^{n+1}}{5^{n+1+1}\cdot(n+1)} \\ a_{n+1}=\frac{2^{n+1}}{5^{n+2}\cdot(n+1)} \end{gathered}[/tex]
Then r is calculated as follows:
[tex]\begin{gathered} r=\lim _{n\to\infty}|\frac{a_{n+1}}{a_n}| \\ r=\lim _{n\to\infty}|\frac{2^{n+1}}{5^{n+2}\cdot(n+1)}\text{ / }\frac{2^n}{5^{n+1}\cdot n} \\ r=\lim _{n\to\infty}|\frac{2^{n+1}}{5^{n+2}\cdot(n+1)}\times\frac{5^{n+1}\cdot n}{2^n}| \end{gathered}[/tex]
Multiplying
[tex]\begin{gathered} r=\lim _{n\to\infty}|\frac{2^{n+1}\cdot5^{n+1}\cdot n}{5^{n+2}\cdot(n+1)\cdot2^n} \\ r=\lim _{n\to\infty}|\frac{2^{n+1}}{2^n}\cdot\frac{5^{n+1}}{5^{n+2}}\cdot\frac{n}{n+1}| \\ r=\lim _{n\to\infty}|2^{n+1-n}\cdot5^{n+1-(n+2)}\cdot\frac{n}{n+1}| \end{gathered}[/tex]
Simplify
[tex]r=\lim _{n\to\infty}|2^1\cdot5^{-1}\cdot\frac{n}{n+1}|[/tex]
Apply exponential properties
[tex]\begin{gathered} r=\lim _{n\to\infty}|\frac{2}{5}\cdot\frac{n}{n+1}| \\ r=\lim _{n\to\infty}|\frac{2n}{5\cdot(n+1)}| \\ r=\lim _{n\to\infty}|\frac{2n}{5n+5}| \end{gathered}[/tex]
Applying the limit, the solution is
[tex]r=\frac{2}{5}=0.4[/tex]
So, r = 0.4 and since r is less than 1, the series converges.
Answer: From the ratio test, r = 0.4. The series converges.
