Question

∑_(n=1)^∞▒〖( 1/2 )〗^2n

Answer 1

Answer:

The series converges to $\dfrac{1}{3}$

Step-by-step explanation:

It seems to be this series:

$\sum_{n=1}^{\infty} \left(\dfrac{1}{2} \right)^{2n}$

We have

$\sum_{n=1}^{\infty} \left(\dfrac{1}{2} \right)^{2n} = \sum_{n=1}^{\infty} \left(\dfrac{1}{4} \right)^{n}$

Using the Root test we can see that this series converges once

$\lim_{n \to \infty} \sqrt[n]{|a_n|} < 1 \implies \sum_{n=1}^{\infty} a_n \text{ is convergent}$

Then, $\lim_{n \to \infty} \sqrt[n]{\left(\dfrac{1}{4} \right)^{n}} = \lim_{n \to \infty} \dfrac{1}{4} = \dfrac{1}{4} < 1$

The series is convergent.

Once the series is geometric, the first term is $\dfrac{1}{4}$ and the ratio is also $\dfrac{1}{4}$ in this case.

The sum of infinite geometric series is $S = \dfrac{a_1}{1-r}$ such that $r < 1$

$\therefore S = \dfrac{\frac{1}{4} }{1-\frac{1}{4}} = \dfrac{1}{3}$