convergence of $\sum_{n=1}^\infty\frac{1}{x^{\ln n}}$.

Question

I have the series $$\sum_{n=1}^\infty\dfrac{1}{x^{\ln n}}$$ and I wanna know for what x this series is converge. I think with condensation test if $$\sum_{n=1}^\infty2^n\dfrac{1}{x^{n\ln2}}=\sum_{n=1}^\infty(\dfrac{2}{x^{\ln2}})^n$$ be converged, then we should have $|\dfrac{2}{x^{\ln2}}|<1$ with geometric series, then for $x>\sqrt[\ln2]{2}$ our series is converge.

Is it correct? I appreciate any suggestion or any other solution?

Shashi · Accepted Answer · 2018-01-09T19:06:24.387

2

Your reasoning is correct. It can be done without invoking the condensation test.

Notice that $x\geq 0$ otherwise $x^{\ln(n)}$ is not well defined as a real number. $$x^{\ln(n)}=e^{\ln(x)\ln(n)}=n^{\ln(x)}$$ So you must have $\ln(x)>1$ for convergence. So $x>e$.

You may ask $e\stackrel{?}{=} \sqrt[\ln(2)]{2}$? The answer is yes, because: $$\sqrt[\ln(2)]{2}=2^{1/\ln(2)}=e^{\ln(2)/\ln(2)}=e^1=e$$

edited Jan 09 '18 at 19:06

answered Jan 09 '18 at 18:55

Shashi

8,738

Thanks. Is my reasoning true? – Jan 09 '18 at 18:59
@maryam Yes your reasoning is correct. – Shashi Jan 09 '18 at 19:06
Thanks again . . – Jan 09 '18 at 19:10

convergence of $\sum_{n=1}^\infty\frac{1}{x^{\ln n}}$.

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