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The series $$\sum_{n=2}^{\infty} a^{\log_en}$$ converges for what values of $a$?

Attempt:

I am trying to use the Raabe's test / Logarithmic test .

Let $u_n = a^{\log_en} \implies \dfrac {u_{n}}{u_n+1}= \dfrac { a^{\log_e n } }{a^{\log_e{n+1}}}$

$\implies \dfrac {u_{n}}{u_n+1} = a^{\log_e n -\log_e{n+1}} = a^{ - \log_e(1 + \dfrac {1}{n}) }$

$\implies \dfrac {u_{n}}{u_n+1} = a^{- ( \dfrac {1}{n} - \dfrac {1}{2n^2} + \cdots)}$

$\implies n \bigg( \dfrac {u_{n}}{u_n+1} - 1 \bigg ) = n \bigg( a^{- ( \dfrac {1}{n} - \dfrac {1}{2n^2} + \cdots)} -1 \bigg )$

I am not sure if this method would work ahead for me.

Could someone please give me a way to move forward with regard to this problem/

Thank you very much for your help.

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Observe that $$ a^{\log n} = \exp(\log n \log a) = n^{\log a}. $$ Now, what can you say about the series $\sum_{n=2}^\infty n^{\log a}$?

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  • $\begingroup$ Got it. $a \in (0, \dfrac {1} {e} )$ . Thank you for your answer. $\endgroup$ – MathMan May 10 '15 at 16:41

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