Question

Show that the series summation(1/2^n) converges to the sum 1?
How to find it?

Answer 1

First of all, if you will notice, it is Geometric Series, with-
(1) First Term a = 1 / 2
(2) Common ratio r = 1 / 2

Let's represent this as $G(a = \frac{1}{2}, r = \frac{1}{2})$

$Sum(G)-of-first-n-terms = \frac{a(1 - r^{n} )}{(1 - r)} = \frac{ \frac{1}{2} (1 - ( \frac{1}{2} )^{n} )}{(1 - \frac{1}{2} )} = \frac{ \frac{1}{2} (1 - ( \frac{1}{2} )^{n} )}{(\frac{1}{2} )} \\ \\ = 1 - ( \frac{1}{2} )^{n}$

Now if $\lim_{n \to \infty} Sum(G)(n)$ is calculated, it can be determined whether series converges or diverges. So, finding the value

$\lim_{n \to \infty} (1 - ( \frac{1}{2} )^{n}) \\ \\ \lim_{n \to \infty} (1 - \frac{1}{ 2^{n} }) \\ \\ 1 - ( \lim_{n \to \infty} \frac{1}{ 2^{n} }) \\ \\ 1 - \frac{1}{ 2^{\infty} } \\ \\ 1 - \frac{1}{\infty} \\ \\ 1 - 0 = 1$

Thus
Series Converges to 1

Answer 2

Answer:

Step-by-step explanation:

First of all, if you will notice, it is Geometric Series, with-

(1) First Term a = 1 / 2

(2) Common ratio r = 1 / 2

Let's represent this as 

Now if  is calculated, it can be determined whether series converges or diverges. So, finding the value

Thus

Series Converges to 1