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How do you find a power series representation for $f(x)=\dfrac{1}{1+x}$ and what is the radius of convergence?

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: Power series is a series or sum of a sequence involving powers. The number of elements in the series can be finite or infinite. To represent the given function in the form of a power series take the help of infinite geometric series when $|r|<1$.

Complete step by step solution:
Many mathematical functions can be expressed or represented in the form of power series.Power series is a series or sum of a sequence involving powers. The number of elements in the series can be finite or infinite.A power series can be considered as a function of some variable (say x) . Suppose we have an infinite geometric series,
$S=1+r+{{r}^{2}}+{{r}^{3}}+.....$
This series can expressed with summation notation as,
$S=\sum\limits_{n=0}^{\infty }{{{r}^{n}}}$
We know that the above geometric series converges to $\dfrac{1}{1-r}$ when $|r|<1$. This means that when $|r|<1$,
$\sum\limits_{n=0}^{\infty }{{{r}^{n}}}=\dfrac{1}{1-r}$ ….. (i)
Now, let us consider the expression $\dfrac{1}{1-r}$ as a function by replacing r with x. Then equation (i) changes to $\dfrac{1}{1-x}=\sum\limits_{n=0}^{\infty }{{{x}^{n}}}$.
This means that $\dfrac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+.....$ …. (ii)
Therefore, we found a function that can express or represent a power series and we also know that the above series is a converging series.Now, if we substitute the x as (-x) in equation (ii), then the equation will change in to
$\dfrac{1}{1-(-x)}=1+(-x)+{{(-x)}^{2}}+{{(-x)}^{3}}+.....$
$\Rightarrow \dfrac{1}{1+x}=1-x+{{x}^{2}}-{{x}^{3}}+{{x}^{4}}+.....$
Therefore, we represented the function $f(x)=\dfrac{1}{1+x}$ in the form of power series.If a converging series converges only when $|x|
Note: Some students may get confused between a converging series and a diverging series.A converging series is a series that has a finite value. Whereas a diverging series a series that does not have a finite value.Therefore, when $|r|>1$ the geometric series is a diverging series since it does not have a finite value.