Test the following series for convergence.
Σ from n=1 to ♾️ for n.x^n-1 , x>0
R=1limn→∞nn=11=1;x=−1:∑n=1∞n(−1)ndiverges because limn→∞n(−1)n=∞;x=1:∑n=1∞n⋅1ndiverges because limn→∞n=∞;x∈(−1,1)−interval of convergenceR=\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\,\,convergenceR=n→∞limnn1=11=1;x=−1:∑n=1∞n(−1)ndivergesbecausen→∞limn(−1)n=∞;x=1:∑n=1∞n⋅1ndivergesbecausen→∞limn=∞;x∈(−1,1)−intervalofconvergence
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