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[tex] \displaystyle \sum_{n = 1}^{ \infty } {4}^{ - n} [/tex]
evaluation of the sum above ​

Sagot :

semsee45

Answer:

[tex]\dfrac{1}{3}[/tex]

Step-by-step explanation:

Given:

[tex]\displaystyle \sum^{\infty}_{n=1} 4^{-n}[/tex]

The sigma notation means to find the sum of the given geometric series where the first term is when n = 1 and the last term is when n = ∞.

To find the first term in the series, substitute n = 1 into the expression:

[tex]\implies a_1=4^{-1}=\dfrac{1}{4}[/tex]

The common ratio of the geometric series can be found by dividing one term by the previous term:

[tex]\implies r=\dfrac{a_2}{a_1}=\dfrac{4^{-2}}{4^{-1}}=\dfrac{1}{4}[/tex]

As | r | <  1  the series is convergent.

When a series is convergent, we can find its sum to infinity (the limit of the series).

Sum to infinity formula:

[tex]S_{\infty}=\dfrac{a}{1-r}[/tex]

where:

  • a is the first term in the series
  • r is the common ratio

Substitute the found values of a and r into the formula:

[tex]S_{\infty}=\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac{1}{3}[/tex]

Therefore:

[tex]\displaystyle \sum^{\infty}_{n=1} 4^{-n}=\dfrac{1}{3}[/tex]

Learn more about Geometric Series here:

https://brainly.com/question/27737241

https://brainly.com/question/27948054

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