I have this math problem that states:
In each part, write out the first four terms of the series, and then find the radius of convergence. $$(a) \sum_{k=1}^{\infty}{\frac{1\cdot2\cdot3\cdot\cdot\cdot k}{1\cdot 6\cdot 11\cdot\cdot\cdot (5k-4)}}x^k$$ $$(b) \sum_{k=1}^{\infty}{(-1)^k\frac{1\cdot 2\cdot 3\cdot\cdot\cdot k}{1\cdot 8\cdot 15\cdot\cdot\cdot (7k-6)}}x^{2k+1}$$
For $(a)$ I took the $a_n$ (which was $\frac{k}{(5k-4)}$) as plugged in 1-4 and got $1+\frac{1}{3}+\frac{3}{11}+\frac{1}{4}$.
I then took $\lim_{k->\infty}{\frac{k}{(5k-4)}}$ and got $L=\frac{1}{5}->R=5$
for (b) I took the $a_n$ (which was $\frac{k}{(7k-6)}$) as plugged in 1-4 and got $1+\frac{1}{4}+\frac{1}{5}+\frac{2}{11}$.
I then took $\lim_{k->\infty}{\frac{k}{(7k-6)}}$ and got $L=\frac{1}{7}->R=7$
I'm not sure what I did wrong. Thanks.
Remember that $a_k$ refers to the $k$:th term. That is, in (b), $$a_k = (-1)^k\frac{1\cdots k}{1\cdot 8 \cdots (7k-6)} x^{2k+1}.$$ So, we find that $$\left|\frac{a_{k+1}}{a_k}\right| = \frac{k+1}{7k+1}|x|^2 \to \frac{|x|^2}{7}.$$ Here, the right hand side is less than $1$ if and only if $|x| < \sqrt{7}$.