Question #314403

Test the following series for convergence.


Σ from n=1 to ♾️ for n.x^n-1 , x>0

1
Expert's answer
2022-03-23T04:09:59-0400

R=1limnnn=11=1;x=1:n=1n(1)ndivergesbecauselimnn(1)n=;x=1:n=1n1ndivergesbecauselimnn=;x(1,1)intervalofconvergenceR=\frac{1}{\underset{n\rightarrow \infty}{\lim}\sqrt[n]{n}}=\frac{1}{1}=1;\\x=-1:\sum_{n=1}^{\infty}{n\left( -1 \right) ^n}diverges\,\,because\,\,\underset{n\rightarrow \infty}{\lim}n\left( -1 \right) ^n=\infty; \\x=1:\sum_{n=1}^{\infty}{n\cdot 1^n}diverges\,\,because\,\,\underset{n\rightarrow \infty}{\lim}n=\infty; \\x\in \left( -1,1 \right) -interval\,\,of\,\,convergence


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