<s>[INST] Interval of convergence homework $$ \sum_{n=1}^{\infty} \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} (x-1)^n $$
First, I start with the ratio test:
$$\lim_{n\to \infty} \left| \left(\frac{-1}{4}\right)^{n+1} \frac{(5(n+1))^{n+1}}{(n+1)!} (x-1)^{n+1} \div \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} (x-1)^n \right| =$$
$$=\frac{1}{4} \lim_{n\to \infty} \left| \frac{(5(n+1))^{n+1} (x-1)^{n+1} n!}{(n+1)!(5n)^n (x-1)^n}\right|= \left| \frac{x-1}{4}\right| \lim_{n\to \infty} \left| \frac{5(n+1)^{n+1}}{n^{n+1}} \right|=$$
$$= \left| \frac{5x-5}{4}\right|\lim_{n\to \infty} \left| \frac{(n+1)^{n+1}}{n^{n+1}} \right|=\left| \frac{5ex-5e}{4}\right|$$
For the series to converge:
$$-\frac{5}{4e} +1 < x < \frac{5}{4e}+1$$
Now I should check the endpoints:
$$ \sum_{n=1}^{\infty} \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} \left(-\frac{5}{4e}\right)^n$$
and $$\sum_{n=1}^{\infty} \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} \left(\frac{5}{4e}\right)^n$$
After checking my answers with wolfram, I think I've made a mistake somewhere. Can somebody help me out.
 [/INST] You made a small algebra error in solving the inequality at the end, interchanging the $4$ and the $5$: you should have
$$1-\frac4{5e}<x<1+\frac4{5e}\;.$$
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