state and prove cauchy root test
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The proof of the convergence of a series Σanis an application of the comparison test. If for all n ≥ N (N some fixed natural number) we have {\displaystyle {\sqrt[{n}]{|a_{n}|}}\leq k<1,} then {\displaystyle |a_{n}|\leq k^{n}<1}. Since the geometric series {\displaystyle \sum _{n=N}^{\infty }k^{n}} converges so does {\displaystyle \sum _{n=N}^{\infty }|a_{n}|} by the comparison test. Hence Σan converges absolutely. Note that {\displaystyle \limsup _{n\to \infty }a_{n}\leq C<1} implies that {\displaystyle |a_{n}|\leq {\frac {C+1}{2}}<1} for almost all {\displaystyle n\in \mathbb {N} }.
If 1}">{\displaystyle {\sqrt[{n}]{|a_{n}|}}>1} 1" style="margin: 0px; padding: 0px; border: 0px; font-style: inherit; font-variant: inherit; font-weight: inherit; font-stretch: inherit; line-height: inherit; font-family: inherit; font-size: 16px; vertical-align: middle; background: none; display: inline-block; animation: fadeInImage 0.3s ease-in 0s 1 normal none running; width: 10.327ex; height: 4.843ex;"> for infinitely many n, then anfails to converge to 0, hence the series is divergent.
Proof of corollary: For a power series Σan = Σcn(z − p)n, we see by the above that the series converges if there exists an N such that for all n ≥ N we have
{\displaystyle {\sqrt[{n}]{|a_{n}|}}={\sqrt[{n}]{|c_{n}(z-p)^{n}|}}<1,}
equivalent to
{\displaystyle {\sqrt[{n}]{|c_{n}|}}\cdot |z-p|<1}
for all n ≥ N, which implies that in order for the series to converge we must have {\displaystyle |z-p|<1/{\sqrt[{n}]{|c_{n}|}}} for all sufficiently large n. This is equivalent to saying
{\displaystyle |z-p|<1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}},}
so {\displaystyle R\leq 1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.} Now the only other place where convergence is possible is when
{\displaystyle {\sqrt[{n}]{|a_{n}|}}={\sqrt[{n}]{|c_{n}(z-p)^{n}|}}=1,}
(since points > 1 will diverge) and this will not change the radius of convergence since these are just the points lying on the boundary of the interval or disc, so
{\displaystyle R=1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.}
If 1}">{\displaystyle {\sqrt[{n}]{|a_{n}|}}>1} 1" style="margin: 0px; padding: 0px; border: 0px; font-style: inherit; font-variant: inherit; font-weight: inherit; font-stretch: inherit; line-height: inherit; font-family: inherit; font-size: 16px; vertical-align: middle; background: none; display: inline-block; animation: fadeInImage 0.3s ease-in 0s 1 normal none running; width: 10.327ex; height: 4.843ex;"> for infinitely many n, then anfails to converge to 0, hence the series is divergent.
Proof of corollary: For a power series Σan = Σcn(z − p)n, we see by the above that the series converges if there exists an N such that for all n ≥ N we have
{\displaystyle {\sqrt[{n}]{|a_{n}|}}={\sqrt[{n}]{|c_{n}(z-p)^{n}|}}<1,}
equivalent to
{\displaystyle {\sqrt[{n}]{|c_{n}|}}\cdot |z-p|<1}
for all n ≥ N, which implies that in order for the series to converge we must have {\displaystyle |z-p|<1/{\sqrt[{n}]{|c_{n}|}}} for all sufficiently large n. This is equivalent to saying
{\displaystyle |z-p|<1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}},}
so {\displaystyle R\leq 1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.} Now the only other place where convergence is possible is when
{\displaystyle {\sqrt[{n}]{|a_{n}|}}={\sqrt[{n}]{|c_{n}(z-p)^{n}|}}=1,}
(since points > 1 will diverge) and this will not change the radius of convergence since these are just the points lying on the boundary of the interval or disc, so
{\displaystyle R=1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.}
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In mathematics, the root test is a criterion for the convergence (a convergence test) of an infinite series. It depends on the quantity
{\displaystyle \limsup _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}},}
where {\displaystyle a_{n}} are the terms of the series, and states that the series converges absolutely if this quantity is less than one but diverges if it is greater than one. It is particularly useful in connection with power series.
{\displaystyle \limsup _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}},}
where {\displaystyle a_{n}} are the terms of the series, and states that the series converges absolutely if this quantity is less than one but diverges if it is greater than one. It is particularly useful in connection with power series.
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