I need to use the ratio test to determine if the infinite series $\displaystyle \frac{3^n}{2^n +1}$ converges or diverges.
I began with $\frac{a_{n+1}}{a_n}$ and got to $\frac{3(2^n+1)}{2^{n+1}+1}$ but I don't really know where to go next...
I need to use the ratio test to determine if the infinite series $\displaystyle \frac{3^n}{2^n +1}$ converges or diverges.
I began with $\frac{a_{n+1}}{a_n}$ and got to $\frac{3(2^n+1)}{2^{n+1}+1}$ but I don't really know where to go next...
You can also do this which is far simpler.
${3^n\over 2^n + 1} = \left({3\over 2}\right)^n{1\over{1 + 1/2^n}}.$ The second factor converges to 1. The first increases without bound. This diverges.
You would only use the ratio test if you were interested in the convergence of the series
$\sum_n {3^n\over 1+ 2^n}$
This would diverge too since the terms do not go to zero.
you came to
$\frac{3(2^n+1)}{2^{n+1}+1}$
divide numerator and denominator with ${2^n}$.
$\frac{3(1+\frac{1}{2^n})}{2+\frac{1}{2^{n}}}$
take the limit of $n$ to infinity.
$\lim_{n\to\infty}\frac{3(1+\frac{1}{2^n})}{2+\frac{1}{2^{n}}}=\frac{3}{2}$
this has to diverge, since
$\frac{3}{2}>1$