Use a power series to test the series ∑n=1∞nlog⁡x for convergence, where ∑ means summation

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Maciej Morrow · Accepted Answer

We obtain the assemblage of∑n=1∞nlog⁡xIt is now necessary to confirm the convergence of the aforementioned seriesHere,un=nlog⁡xBy ratio test,limn→∞un+1un=limn→∞(n+1)log⁡xnlog⁡x=limn→∞(n+1n)log⁡x=limn→∞(1+1n)log⁡x=1The test is therefore inconclusive.Now, by root test,=limn→∞un1n=limn→∞(nlog⁡x)1n=limn→∞((n)1n)log⁡x=(limn→∞n1n)log⁡x=1So, the test is inconclusive.Now, we&#039;ll use a comparison test to try to find the convergenceLet us consider the series,&amp;nbsp;∑n=0∞1npThe above is called the p-series and it converges if and only if p&amp;gt;1 or -p&amp;lt;-1Now, comparing&amp;nbsp;an=1np=n−p&amp;nbsp;and&amp;nbsp;un=nlog⁡xThen the series&amp;nbsp;un&amp;nbsp;converges if and only if&amp;nbsp;log⁡x&amp;lt;−1And&amp;nbsp;log⁡x&amp;lt;−1⇒x&amp;lt;10−1⇒x∈(0,110)Thus, the given series converges for&amp;nbsp;x∈(0,110)

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Use a power series to test the series sum_{n=1}^infty n^{log x} for convergence, where sum means summation

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