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Question: For $a>0$. the series $$ \sum_{n=2}^{\infty} a^{\log_e n}$$

is convergent for which range of values of $a$?

My Approach:

The divergence test of checking $\lim_{n\to\infty} t_n =0 $ gives $0<a<1$

So does the ratio test and the root test.

I am fairly new to this topic, so I am not sure how to proceed.

Edit: The answer is $0<a<1/e$ I fail to see how they arrive at this.

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1 Answer 1

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Hint:

This is indeed a Riemann series, since you can rewrite its general term as $$a^{\log n}=\mathrm e^{\log n\log a}=n^{\log a}.$$ As $\sum_n n^\alpha$ converges if and only if $\alpha <-1$, this yields $$\log a<-1\iff 0<a<\dfrac1{\mathrm{e}}.$$

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  • $\begingroup$ I looked up Riemann series, but could not make the jump. Could you elaborate please? $\endgroup$ Commented May 11, 2018 at 17:55
  • $\begingroup$ I have edited question details. $\endgroup$ Commented May 11, 2018 at 17:57
  • $\begingroup$ Is it clearer now? $\endgroup$
    – Bernard
    Commented May 11, 2018 at 18:12
  • $\begingroup$ (+1) Yes, it is. Thanks. $\endgroup$ Commented May 11, 2018 at 18:14

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