Question

HelloI’m having trouble on the *calculus* portion on my ACT PREP GUIDEI need help solving this

Answer 1

First, we find the value of r:

`\begin{gathered} a_n=(2^n)/(5^(n+1)\cdot n) \\ a_(n+1)=(2^(n+1))/(5^(n+1+1)\cdot(n+1)) \\ a_(n+1)=(2^(n+1))/(5^(n+2)\cdot(n+1)) \end{gathered}`

Then r is calculated as follows:

`\begin{gathered} r=\lim _(n\to\infty)|(a_(n+1))/(a_n)| \\ r=\lim _(n\to\infty)|(2^(n+1))/(5^(n+2)\cdot(n+1))\text{ / }(2^n)/(5^(n+1)\cdot n) \\ r=\lim _(n\to\infty)|(2^(n+1))/(5^(n+2)\cdot(n+1))*(5^(n+1)\cdot n)/(2^n)| \end{gathered}`

Multiplying

`\begin{gathered} r=\lim _(n\to\infty)|(2^(n+1)\cdot5^(n+1)\cdot n)/(5^(n+2)\cdot(n+1)\cdot2^n) \\ r=\lim _(n\to\infty)|(2^(n+1))/(2^n)\cdot(5^(n+1))/(5^(n+2))\cdot(n)/(n+1)| \\ r=\lim _(n\to\infty)|2^(n+1-n)\cdot5^(n+1-(n+2))\cdot(n)/(n+1)| \end{gathered}`

Simplify

`r=\lim _(n\to\infty)|2^1\cdot5^(-1)\cdot(n)/(n+1)|`

Apply exponential properties

`\begin{gathered} r=\lim _(n\to\infty)|(2)/(5)\cdot(n)/(n+1)| \\ r=\lim _(n\to\infty)|(2n)/(5\cdot(n+1))| \\ r=\lim _(n\to\infty)|(2n)/(5n+5)| \end{gathered}`

Applying the limit, the solution is

`r=(2)/(5)=0.4`

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.