Question

evaluate sum from n=1 to infinity ((3n-1)/(4n+3))^(2n)

by the root test, is it divergent, convergent or inconclusive?

Answer 1

converges for sure. that \(^{2n}\) in the denominator makes the terms go to zero lickety split

Answer 2

\[\lim_{n\to \infty}\left(\frac{3n-1}{(4n+3)^{2n}}\right)^{\frac{1}{n}}\]
\[\lim_{n\to \infty}\left(\frac{(3n+1)}{(4n+3)^2}\right)=0\]

Answer 3

oops i made a mistake, should be a \(\frac{1}{n}\) in the numerator as well

Answer 4

the answer was 9/16...

Answer 5

\[
\left(\left(\frac{3 n-1}{4
n+3}\right)^{2
n}\right)^{\frac{1}{n}}=\frac{(3 n-1)^2}{(4 n+3)^2}
\]
For n near infinity the above ration behaves like
\[
\frac { 9n^2}{16 n^2}= \frac 9 {16}
\]

Answer 6

To see it in more steps,
\[
\frac{(3 n-1)^2}{(4 n+3)^2}= \frac {9 n^2-6 n+1 }
{16 n^2+24 n+9} = \frac { \frac {9 n^2-6 n+1 } {n^2} }
{\frac{16 n^2+24 n+9}{n^2}}=\frac { 9 - \frac 6 n + \frac 1 {n^2}}
{16 + \frac {24}{n}+ \frac 9{n^2}
}
\]
you can see that the ratio tends to 9/16 when n goes to infinity.

Answer 7

Is it convergent or divergent?