Question

Pls i literally do not know how to do this

Answer 1

Answer:

• r = (n+1)/4
• the series diverges

Step-by-step explanation:

The ratio test asks you to look at the ratio of two successive terms of the sum. It is generally convenient to look at the ratio ...

  $\dfrac{a_{n+1}}{a_n}$

where $a_n$ is the term shown in parentheses to the right of the summation symbol. Here, that ratio is ...

  $\dfrac{\left(\dfrac{2(n+1)!}{2^{2(n+1)}}\right)}{\left(\dfrac{2n!}{2^{2n}}\right)}=\dfrac{2(n+1)!\cdot 2^{2n}}{2n!\cdot 2^{2(n+1)}}\\\\=\dfrac{2(n+1)!}{2n!}\cdot\dfrac{2^{2n}}{2^{2(n+1)}}=\dfrac{n+1}{(2^2)} = \dfrac{n+1}{4}$

Most of the terms of the factorial product cancel, and the powers of 2 all cancel except for 2^2. So, the ratio of adjacent terms is ...

  $\boxed{r_n=\dfrac{n+1}{4}}$

This gets larger and larger as n gets larger, so the series diverges.