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

HelloI’m having trouble on the *calculus* portion on my ACT PREP GUIDEI need help-example-1

1 Answer

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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.

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