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∑_(n=1)^∞▒〖( 1/2 )〗^2n

User Fvillalba
by
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1 Answer

4 votes

Answer:

The series converges to
(1)/(3)

Explanation:

It seems to be this series:


$ \sum_(n=1)^(\infty) \left((1)/(2) \right)^(2n)$

We have


$ \sum_(n=1)^(\infty) \left((1)/(2) \right)^(2n) = \sum_(n=1)^(\infty) \left((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,
$\lim_(n \to \infty) \sqrt[n]{\left((1)/(4) \right)^(n)} = \lim_(n \to \infty) (1)/(4) = (1)/(4) < 1$

The series is convergent.

Once the series is geometric, the first term is
(1)/(4) and the ratio is also
(1)/(4) in this case.

The sum of infinite geometric series is
S = (a_1)/(1-r) such that
r < 1


\therefore S = ((1)/(4) )/(1-(1)/(4)) = (1)/(3)

User Tokfrans
by
6.2k points