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Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.4
Determine the radius of convergence and the interval of convergence for the power series $\sum _{n\=1}^{\infty}n{x}^{n}$.
Even though (7) in Table 8.4.1 claims that absolute convergence at one end of the interval of convergence implies absolute convergence at the other, if the convergence at an endpoint is absolute, verify that it also absolute at the other.
Solution
Since the given power series contains the powers ${x}^{n}$, the radius of convergence is given by
$R\=\underset{n\to \infty}{lim}{a}_{n}\/{a}_{n\+1}$ = $\underset{n\to \infty}{lim}\frac{n}{n\+1}\=1$
At the right endpoint $x\=R\=1$, the given power series becomes $\mathrm{\Σ}n$, which diverges by the nth-term test.
At the left endpoint $x\=-R\=-1$, the given power series becomes the alternating series $\mathrm{\Σ}{\left(-1\right)}^{n}n$, which does not yield to the Leibniz test because $n$ does not tend to zero. However, for that very reason, it does yield to the nth-term test, which indicates that the series diverges.
Hence, the interval of convergence is $\left(-R\,R\right)\=\left(-1\,1\right)$.
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