Does the converge or diverge? sum_(n=1)^(oo)(1)/(2^(n))=(1)/(2)+(1)/(4)+(1)/(8)+(1)/(16)+cdots

Given series: We are given the infinite geometric series $ \sum_{n=1}^{\infty} \frac{1}{2^n} $. To find the sum of an infinite geometric series, we use the formula $ S = \frac{a}{1 - r} $, where $ a $ is the first term and $ r $ is the common ratio. In our case, the first term $ a = \frac{1}{2} $ and the common ratio $ r = \frac{1}{2} $.Formula application: Now we apply the formula for the sum of an infinite geometric series: $ S = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} $.Simplify expression: Simplify the expression: $ S = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 $.Final result: The sum of the infinite series $ \sum_{n=1}^{\infty} \frac{1}{2^n} $ is 1.

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Does the converge or diverge?
sum_(n=1)^(oo)(1)/(2^(n))=(1)/(2)+(1)/(4)+(1)/(8)+(1)/(16)+cdots

Does the series converge or diverge? $\sum_{n=1}^{\infty}\frac{1}{2^{n}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$

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Calculus

Evaluate definite integrals using the chain rule

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Q. Does the series converge or diverge?

\sum_{n=1}^{\infty}\frac{1}{2^{n}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots

Given series: We are given the infinite geometric series $\sum_{n=1}^{\infty} \frac{1}{2^n}$ . To find the sum of an infinite geometric series, we use the formula $S = \frac{a}{1 - r}$ , where $a$ is the first term and $r$ is the common ratio. $\newline$ In our case, the first term $a = \frac{1}{2}$ and the common ratio $r = \frac{1}{2}$ .
Formula application: Now we apply the formula for the sum of an infinite geometric series: $\newline$ $S = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \frac{1}{2}}$ .
Simplify expression: Simplify the expression: $\newline$ $S = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$ .
Final result: The sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{2^n}$ is $1$ .