For the ratio test of a series the ratio is defined as:
[tex]r=\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert[/tex]And we have three possible outcomes:
- r<1 and the series converges.
- r>1 and the series diverges.
- r=1 and the test is inconclusive.
Knowing this let's apply the test to the series given by the question. We have:
[tex]\begin{gathered} a_n=\frac{2n!}{2^{2n}} \\ a_{n+1}=\frac{2(n+1)!}{2^{2(n+1)}} \end{gathered}[/tex]As you can see both expressions are always positive so when writing the ratio we don't need to add the absolute value symbols. Then the ratio is:
[tex]\begin{gathered} r=\frac{a_{n+1}}{a_n}=\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}=\frac{2(n+1)!}{2^{2(n+1)}}\cdot\frac{2^{2n}}{2n!} \\ r=\frac{2(n+1)!}{2n!}\cdot\frac{2^{2n}}{2^{2n+2}} \end{gathered}[/tex]Here is important to remember a property of the factorial:
[tex]\begin{gathered} a!=1\cdot2\cdot3\cdot\ldots\cdot(a-1)\cdot a \\ (a+1)!=1\cdot2\cdot3\cdot\ldots\cdot(a-1)\cdot a\cdot(a+1) \\ (a+1)!=a!\cdot(a+1) \end{gathered}[/tex]And a property of powers:
[tex]b^{a+c}=b^a\cdot b^c[/tex]Using these properties we get:
[tex]\begin{gathered} r=\frac{2(n+1)!}{2n!}\cdot\frac{2^{2n}}{2^{2n+2}}=\frac{2n!\cdot(n+1)}{2n!}\cdot\frac{2^{2n}}{2^{2n}\cdot2^2} \\ r=\frac{2n!\cdot(n+1)}{2n!}\cdot\frac{2^{2n}}{2^{2n}\cdot2^2}=\frac{2n!}{2n!}\cdot(n+1)\cdot\frac{2^{2n}}{2^{2n}}\cdot\frac{1}{2^2} \\ r=\frac{n+1}{2^2}=\frac{n+1}{4} \\ r=\frac{n+1}{4} \end{gathered}[/tex]So the answer to part (a) is:
[tex]r=\frac{n+1}{4}[/tex]For part (b) we have:
The ratio depends on n. As n increases the ratio increases and gets greater than 1:
[tex]\begin{gathered} r>1 \\ \frac{n+1}{4}>1 \\ n+1>4 \\ n>3 \end{gathered}[/tex]So for any n>3 the ratio is greater than 1 which means that this value of r tells us that the series diverges.
