Integrate the power series for \sum \limits_ {n=1}^{\infty} \frac{8}{(1+x)^n} at x=0 to obtain a...

Question:

Integrate the power series for {eq}\sum \limits_ {n=1}^{\infty} \frac{8}{(1+x)^n} {/eq} at {eq}x=0 {/eq} to obtain a power series representation for the function {eq}f(x)= 8 \ln(1+x) {/eq}

Indefinite Integral:

Assume that {eq}v {/eq} is a single-valued function of {eq}x {/eq} and {eq}K {/eq} is an anti derivative of {eq}v {/eq} such that {eq}\displaystyle v\left( x \right) = K'\left( x \right),\;\forall \;x {/eq}.

Then the indefinite integral of {eq}\displaystyle v {/eq} is defined as {eq}\displaystyle \int v\left( x \right)dx = K\left( x \right) + C {/eq}, where {eq}c {/eq} is an integration constant.

Answer and Explanation:

Given that: {eq}\displaystyle \sum\limits_{n = 1}^\infty {\frac{8}{{{{(1 + x)}^n}}}} {/eq}

{eq}\displaystyle \eqalign{ & \sum\limits_{n = 1}^\infty {\frac{8}{{{{(1 + x)}^n}}}} \cr & {\text{From integration;}} \cr & \int {\sum\limits_{n = 1}^\infty {\frac{8}{{{{(1 + x)}^n}}}} dx} \cr & \sum\limits_{n = 1}^\infty 8 \int {{{(1 + x)}^{ - n}}dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{a^{ - n}} = \frac{1}{{{a^n}}}} \right) \cr & \sum\limits_{n = 2}^\infty 8 \left( {\frac{{{{(1 + x)}^{1 - n}}}}{{1 - n}} + c} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + c} } \right) \cr & \sum\limits_{n = 2}^\infty {\frac{1}{{1 - n}}} \left( {\frac{8}{{{{(1 + x)}^{n - 1}}}}} \right) + \sum\limits_{n = 2}^\infty {8c} \cr & \cr & {\text{At }}x = 0; \cr & \sum\limits_{n = 1}^\infty {\frac{8}{{{{(1 + 0)}^n}}}} + \sum\limits_{n = 2}^\infty {8c} \cr & \sum\limits_{n = 1}^\infty 8 \cr & {\text{Series is divergent}}{\text{.}} \cr & \cr & f(x) = 8\ln (1 + x) \cr & {\text{At }}x = 0; \cr & f(0) = 8\ln (1 + 0) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 8\left( 0 \right) \cr & \cr & f(0) = 0 \cr} {/eq}


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Indefinite Integral: Definition, Rules & Examples

from Calculus: Tutoring Solution

Chapter 7 / Lesson 14
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