\(1+3+..+3^n=\dfrac{3^{n+1}-1}{3-1}=\dfrac{3^{n+1}-1}{2}\)
\(1+2+...+2^n=\dfrac{2^{n+1}-1}{2-1}=2^{n+1}-1\)
\(I=\lim\limits\dfrac{3^{n+1}-1}{2.2^{n+1}-2}=\lim\dfrac{\left(\dfrac{3}{2}\right)^{n+1}-\left(\dfrac{1}{2}\right)^{n+1}}{2-2\left(\dfrac{1}{2}\right)^{n+1}}=\dfrac{+\infty}{2}=+\infty\)



