Question

I need help with this practice problem Answer these below as well, The value of r from the ratio test is A. 0B. 0.25C. 1D. 4The series A. Converges B. Diverges

Answer 1

Given the series below

$\sum ^{\infty}_{n\mathop=1}\frac{4^{n+1}}{4n+1}$

By the limit comparison test, the series diverges

Hence, the series diverges

Where

$\begin{gathered} a_n=\frac{4^{n+1}}{4n+1} \\ \text{And } \\ a_{n+1}=\frac{4^{n+1+1}}{4(n+1)+1}=\frac{4^{n+2_{}}}{4n+1+1}=\frac{4^{n+2}}{4n+2} \end{gathered}$

For the ratio test,

$=\frac{a_{n+1}}{a_n}$

Substitute for the series

$=(\frac{4^{n+2}}{4n+2})\times(\frac{4n+1}{4^{n+1}})$

Solve

$\begin{gathered} r=\frac{(4^{n+2-(n+1)_{}})(4n+1)_{}}{4n+2_{}} \\ =\frac{(4^{n-n+2-1})(4n+1)}{4n+2} \\ =\frac{4^1(4n+1)}{4n+2} \\ =\frac{16n+4}{4n+2} \\ =\frac{2^2(4n+1)^{}}{2(n+1)} \\ =\frac{2(4n+1)}{2n+1}\text{ } \\ \lim _{n\rightarrow\infty} \\ =2(2)=4 \end{gathered}$

The value of r from the ratio test is 4

Since r is greater than 1 i.e r > 1

Hence, the series diverges