Subtract 5 each play so go backwards from zero and you should finish at -15
The ricipricals are 3/8 and 2/13
It would be 50.24 ft
BTW in this case I used 3.14 for pi
You can use the root test here. The series will converge if
![L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}](https://tex.z-dn.net/?f=L%3D%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cfrac%7B%284-x%29%5En%7D%7B4%5En%2B9%5En%7D%7D%3C1)
You have
![L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}=|4-x|\lim_{n\to\infty}\frac1{\sqrt[n]{4^n+9^n}}](https://tex.z-dn.net/?f=L%3D%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cfrac%7B%284-x%29%5En%7D%7B4%5En%2B9%5En%7D%7D%3D%7C4-x%7C%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac1%7B%5Csqrt%5Bn%5D%7B4%5En%2B9%5En%7D%7D)
Notice that
![\dfrac1{\sqrt[n]{4^n+9^n}}=\dfrac1{\sqrt[n]{9^n}\sqrt[n]{1+\left(\frac49\right)^n}}=\dfrac1{9\sqrt[n]{1+\left(\frac49\right)^n}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%5Bn%5D%7B4%5En%2B9%5En%7D%7D%3D%5Cdfrac1%7B%5Csqrt%5Bn%5D%7B9%5En%7D%5Csqrt%5Bn%5D%7B1%2B%5Cleft%28%5Cfrac49%5Cright%29%5En%7D%7D%3D%5Cdfrac1%7B9%5Csqrt%5Bn%5D%7B1%2B%5Cleft%28%5Cfrac49%5Cright%29%5En%7D%7D)
so as

, you have

, which means you end up with

This is the interval of convergence. The radius of convergence can be determined by finding the half-length of the interval, or by solving the inequality in terms of

so that

is the ROC. You get
I don’t know if I’m right but I got 2/3 but you should use desmos