Vtyjkyuk

I was trying to find the radius of convergence for $sum_k=1^infty k^33^-k!(x-3)^k!$.

First I rewrote the sum to $sum_k=1^infty frack^3(x-3)^k!3^k!$ and used the ratio test: $lim_k to infty|frac(k+1)^3(x-3)^(k+1)!3^k!k^3(x-3)^k!3^(k+1)!|$.

Now we can write this as: $lim_k to infty|frac(k+1)^3(x-3)^(k+1)!-k!3^k!-(k+1)!k^3|$.

Here we can solve $lim_k to infty|frac(k+1)^3k^3| = |frac(k(1+frac1k))^3k^3| = |frack^3(1+frac1k)^3k^3| = 1$.

Thus, we now have to solve: $lim_k to infty|(x-3)^(k+1)!-k!cdot3^k!-(k+1)!|$.

We can further use the fact that $(k+1)!-k! = k cdot k!$ and $k!-(k+1)! = -k cdot k!$.

Hence, $lim_k to infty|(x-3)^kk!cdot3^-kk!| = |(x-3)^kk!cdotfrac13^kk!| = |frac(x-3)^kk!3^kk!| = |(frac(x-3)3)^kk!|$.

Obviously, if $|x-3| > 3$ this will diverge. This happens if $x > 6$ or $x < 0$. That's why I concluded that $L = infty > 1$ if and only if $x neq [0,...,6]$. Therefore, $R = 0$ and Interval of convergence: $0 < x < 6$.

Is this solution completely wrong? It feels like maybe I made a mistake somewhere and Wolfram Alpha couldn't compute this. I'm very happy about every help I can get or answer!

When you calculate the radius of convergence of a power series you can just work with the coefficients; values of $x$ don't come into the argument. Let $a_n=k^33^-k!$ if $n=k!$ and $0$ otherwise. Then the given series is $sum a_n (x-3)^n$. The general formula for radius $R$ of convergence of the series is given by $frac 1 R=lim sup |a_n|^1/n$. Hence $frac 1 R =lim sup_k k^3/k! 3^-1=frac 1 3$. Hence $R=3$. (The domain of convergence is $x-3$ but you have only been asked to find the radius of convergence).