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I have to find for which parameters is series $\sum_{n = 1}^{+\infty} \frac{\alpha^n}{n + \beta^n}, \beta > 0$ convergent.

Trying absolute convergence $|a_n|$ and using Cauchy criterion gives me $$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{\alpha^n}{n + \beta^n}} = \frac{\alpha}{\beta}\frac{1}{\sqrt[n]{\frac{n}{\beta^n} + 1}} \to \frac{\alpha}{\beta} < 1. $$

From this we can see that $\alpha < \beta$. If $\alpha = \beta$ we have $$\frac{\alpha^n}{n + \alpha^n} = \frac{1}{\frac{n}{\alpha^n} + 1}$$ which we can compare with some exponential series $\gamma^n$, which will yield that series is convergent if $\alpha \in (0, 1)$.

So series is absolutely convergent for $\alpha = 0$ or $\alpha > 0$ and $\beta > \alpha$.

For $\alpha < 0$ we can rewrite $a_n$-th term as $\frac{(-1)^n \alpha^n}{n + \beta^n}$ and use Leibniz criterion (since $a_n$ is monotonically decreasing) and $\lim a_n = 0 \iff \alpha < \beta$.

So, totally, series converge for $|\alpha| < \beta$. But my textbook states that answer is $|\alpha| < \max\{1, \beta\}$. Do I have mistake in my reasoning or is it an error in the book?

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  1. $\beta>1$ the general term is equivalent to $$\frac {\alpha^n}{\beta^n} $$ and converges if $|\alpha|<\beta $.

  2. 0 <$\beta\leq 1$ the general term is equivalent to $$\frac {\alpha^n}{n} $$ and converges if $|\alpha|<1$ or $\alpha=-1$.

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  • $\begingroup$ I support this answer, because there are too much blind application of convergence criteria when simple equivalents bring easy solutions.But beware anyway, $\alpha=-1$ case has to be studied aside because equivalent work only for absolute convergence. $\endgroup$ – zwim May 13 '17 at 22:31
  • $\begingroup$ @zwim Yes, the equivalence criterion is the fastest. $\endgroup$ – hamam_Abdallah May 13 '17 at 22:32
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I Assume both $\alpha$ and $\beta$ are positive number. Other cases are similar

First Assume $ 1 \leq \beta $ then $$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{\alpha^n}{n + \beta^n}} = \frac{\alpha}{\beta}\frac{1}{\sqrt[n]{\frac{n}{\beta^n} + 1}} \to \frac{\alpha}{\beta} < 1.$$ and obviously in this case if $\frac{\alpha}{\beta} = 1$ the series is divergent. Therefore the condition becomes $\{(\alpha,\beta) ~ |1 \leq \beta , ~ \alpha < \beta \}$.

Now if $ 0 <\beta < 1$ Then $$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{\alpha^n}{n + \beta^n}} = \frac{\alpha}{\beta}\frac{1}{\sqrt[n]{\frac{n}{\beta^n} + 1}} \to \alpha < 1.$$ and obviously if $\alpha =1$ the series is divergent! So Taking into account the first part the maximal set where the series is convergent is

$$ \{(\alpha,\beta) ~ |1 \leq \beta , ~ \alpha < \beta \} \cup \{(\alpha,\beta) ~ | \beta < 1 , ~ \alpha < 1 \} $$

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