could someone give me an unsolved radius of convergence problem and I will work it and they can critique me?
so follow my work here if you wouldnt mind $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-5)^{n}}{n5^{n}}$...ok so I must find all x such that $\displaystyle \lim_{n \to{\infty}}\bigg|\frac{a_{n+1}}{a_n}\bigg|<1$...so $\displaystyle \lim_{n \to{\infty}}\bigg|\frac{(x-5)^{n+1}}{(n+1)5^{n+1}}\cdot{\frac{n5^{n}}{(x-5)^{n}}}\bigg|<1$..we are left with $\displaystyle \lim_{n \to{\infty}}\bigg|\frac{n(x-5)}{5(n+1)}\bigg|<1$...now evaluating the limit we get $\displaystyle \frac{|x-5|}{5}<1$ so $\displaystyle 0<x<10$...now I must test the endpoints of the test intervals..when I input x=0 we get$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-5)^{n}}{n5^{n}}=\sum_{n=1}^{\infty}\frac{(-1)^{2n+1}5^{n}}{n5^{n}}=\sum_{n=1}^{\infty}\frac{(-1)^{2n+1}}{n}$ and now realizing that for integral numbers $\displaystyle (-1)^{2x+1}=-1$ since $\displaystyle 2n+1$ is a series for all odd integers...so now we have $\displaystyle -\sum_{n=1}^{\infty}\frac{1}{n}$ which is obviously a divergent harmonic series...now when we put 10 in for x after all the simplification we would get$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ and seeing that this is equal to$\displaystyle -\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$ and using the Root test we get that $\displaystyle \lim_{n \to {\infty}}\bigg[\frac{(-1)^{n}}{n}\bigg]^{\frac{1}{n}}=\lim_{n \to{\infty}}\frac{-1}{n^{\frac{1}{n}}}=-1$ and since that is $\displaystyle <1$ it is divergent...so the series only converges on $\displaystyle 0<x<10$...is that right?