Find the set of $x>0$ such that the series $\sum\limits_n x^{\ln{n}}$ converges

If $x>0$, find the set of all values of $x$ such that series is convergent$$\sum_{n=1}^{\infty} x^{\ln{n}}$$

My attempt:- I used Ratio test for finding the set of all values of $x$ such that series is convergent.

$$\lim_{x\to\infty}\frac{x^{\ln{n+1}}}{x^{\ln{n}}}$$

$$=\lim_{x\to\infty}x^{\ln\frac{n+1}{n}}$$ This quantity must be less than one for getting a convergent series, I am not able to judge. Can you please help me to find the interval of convergence?

• Hint: $x^{\ln n}=n^{\ln x}$. – David Mitra Aug 13 '17 at 10:31
• The notion of radius of convergence refers to entire series. The function in your question is not an entire series. You might instead, more properly, ask for the set of values of $x>0$ such that the series converges. – Did Aug 13 '17 at 10:34
• I have edited errors in my question. Please help me to solve. – Unknown x Aug 13 '17 at 10:39
• @RaviPrakash Huh? Certainly not a duplicate. – Did Aug 13 '17 at 12:19
• @user243301 ... $\zeta(-\log x)$ – reuns Aug 13 '17 at 15:35

take $x=e^{-y}$

$$\sum_{n=1}^{\infty}x^{\ln n}=\sum_{n=1}^{\infty}e^{-y\ln n}=\sum_{n=1}^{\infty}\frac{1}{n^y}$$

The last series converges iff $y>1$.

Hence $0<x<\frac{1}{e}$, for which the series converges.

• Why did u made this selection? – Unknown x Aug 13 '17 at 10:54
• you said $x>0$, and I show $\ln$, so I thought I need to find series which do not contain $\ln$.... so I took $x=e^{-y}$. – MAN-MADE Aug 13 '17 at 10:57
• How can I able to say that this is the only interval? – Unknown x Aug 13 '17 at 11:02
• @N.Maneesh note that $x\to y$ is a bijective transformation. So you can not find any other point. – MAN-MADE Aug 13 '17 at 11:05
• @N.Maneesh I don't know where can you find problem like this... I know this competitive exam you are talking abt (NET) has books like previous year question etc. Again I can not suggest you any specific real analysis book,Rudin is good, but not for beginners... I never really followed any specific real analysis book (jump from one book to another actually), also, I studied statistics in your country, so it is not a good idea to ask me actually – MAN-MADE Aug 13 '17 at 13:59

We know $x^{\ln y}=y^{\ln x}$ and here we get $\sum_1 ^\infty x^{\ln n}=\sum_1 ^\infty n^{\ln x}=\sum_1 ^\infty \frac{1}{n^{-\ln x}}$ and thus the series converges if $-\ln x>1$ i.e for $0<x<\frac{1}{e}$

• Why repeat this? – Did Aug 14 '17 at 15:16
• Your answer is just the exact copy of above one. – Unknown x Aug 14 '17 at 15:20