Question

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

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

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