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 \(x\in(0,\frac{1}{10})\)