How can we find the interval of convergence of the following series: $$\sum_{n=1}^{\infty} x^n +\frac{1}{({x^n}{2^n})}.$$
Radius of convergence of sum of two series
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sequences-and-series
algebra-precalculus
2 Answers
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You have two geometric series $$ \sum_{n}x^n $$ which converges for $|x|<1$, and $$ \sum_{n}\frac{1}{2^nx^n}= \sum_{n}\left(\frac{1}{2x}\right)^{\!n} $$ which converges for $$ \left|\frac{1}{2x}\right|<1 $$ that is, $|x|>\frac{1}{2}$.
Intersect.
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Hint: Both expressions are part of the geometric series. the first converges for $|x|<1$ and the second converges for $|x/2|<1$. Use the intersection of both regions to get your radius of convergence.
In order to get this result from the ratio rule:$x^n+x^n/2^n=(1+1/2^n)x^n.$ The coefficient is $(1+1/2^n)$. Now apply the ratio rule.
$R=\lim_{n\infty}|\frac{a_n}{a_{n+1}}|=\lim_{n\infty}|\frac{1+1/2^n}{1+1/2^{n+1}}|=1.$