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For what values of a does the series converge:

$\sum_{n=1}^\infty \frac{1}{(n+n^3)^a}$

?

Following is my thought process:

$\sum_{n=1}^\infty \frac{1}{(n+n^3)^a}$ = $\lim_{k\to \infty}\sum_{n=1}^k \frac{1}{(n+n^3)^a}$ = $\frac{1}{(1+1^3)^a} + \frac{1}{(2+2^3)^a+\frac{1}{(3+3^3)^a}}....$

So I'm thinking as long as $a \ge 1$ it should converge - but I'm not sure.

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    @MichaelBurr But we all know the ratio test fails for p-series like series.... – 2017-01-31

3 Answers 3

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As often, equivalents are the shortest way: the general term is $$\frac{1}{(n+n^3)^a}\sim_\infty\frac{1}{n^{3a}},$$ and the latter converges if and only if $3a>1$, i.e. $\; a>\dfrac13$.

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The denominator is $d_n =(n^3+n)^a =n^{3a}(1+n^{-2})^a $.

$d_n > n^{3a}$ so the sum converges when $3a > 1$ or $a > \frac13$.

$d_n < 2^an^{3a}$ so the sum diverges when $3a \le 1$ or $a \le \frac13$.

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As per the term test, we already know that $a>0$. From there, we recognize the comparison test:

$$\frac1{(n^3+n^3)^a}<\frac1{(n+n^3)^a}<\frac1{(0+n^3)^a}$$

which gives us the p-series.

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    Perhaps one could use $n^3 – 2017-01-31
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    @MichaelBurr I repeat the last comment. (>.<) And thank you for everything XD – 2017-01-31