Question

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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asked 2021-01-08
Use a power series to test the series \(\sum_{n=1}^\infty n^{\log x}\) for convergence, where \(\sum\) means summation

Answers (1)

2021-01-09

We are given the series,
\(\sum_{n=1}^\infty n^{\log x}\)
Now, we need to check the convergence of the above series.
Here,
\(u_n=n^{\log x}\)
By ratio test,
\(\lim_{n\rightarrow\infty}\frac{u_{n+1}}{u_n}=\lim_{n\rightarrow\infty}\frac{(n+1)^{\log x}}{n^{\log x}}\)
\(=\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{\log x}\)
\(=\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^{\log x}\)
\(=1\)
So, the test is inconclusive.
Now, by root test,
\(=\lim_{n\rightarrow\infty}u_n^{\frac1n}=\lim_{n\rightarrow\infty}(n^{\log x})^{\frac1n}\)
\(=\lim_{n\rightarrow\infty}((n)^\frac1n)^{\log x}\)
\(=\left(\lim_{n\rightarrow\infty}n^{\frac1n}\right)^{\log x}\)
\(=1\)
So, the test is inconclusive.
Now, we will try to find the convergence using comparison test.
Let us consider the series, \(\sum_{n=0}^\infty\frac{1}{n^p}\)
The above is called the p-series and it converges if and only if p>1 or -p<-1
Now, comparing \(a_n=\frac{1}{n^p}=n^{-p}\) and \(u_n=n^{\log x}\)
Then the series \(u_n\) converges if and only if \(\log x<-1\)
And \(\log x<-1\Rightarrow x<10^{-1}\)
\(\Rightarrow x\in(0,\frac{1}{10})\)
Thus, the given series converges for \(x\in(0,\frac{1}{10})\)

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