Question

n%7D%20" id="TexFormula1" title=" \displaystyle \sum_{n = 1}^{ \infty } {4}^{ - n} " alt=" \displaystyle \sum_{n = 1}^{ \infty } {4}^{ - n} " align="absmiddle" class="latex-formula">
evaluation of the sum above ​

Answer 1

Answer:

$\dfrac{1}{3}$

Step-by-step explanation:

Given:

$\displaystyle \sum^{\infty}_{n=1} 4^{-n}$

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:

$\implies a_1=4^{-1}=\dfrac{1}{4}$

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

$\implies r=\dfrac{a_2}{a_1}=\dfrac{4^{-2}}{4^{-1}}=\dfrac{1}{4}$

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:

$S_{\infty}=\dfrac{a}{1-r}$

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:

$S_{\infty}=\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac{1}{3}$

Therefore:

$\displaystyle \sum^{\infty}_{n=1} 4^{-n}=\dfrac{1}{3}$

Learn more about Geometric Series here:

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