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$$ \sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2)...(1+x^n)};x>1 $$

Just need a tip on which strategy to use to determine if this series converges or not.

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  • $\begingroup$ try to get an idea of which order the n-th term is. Then compare with simpler series. $\endgroup$
    – Thomas
    Apr 9, 2014 at 17:55
  • $\begingroup$ Would the ratio test work for this question? $\endgroup$
    – user136088
    Apr 9, 2014 at 18:59
  • $\begingroup$ The ratio test does work actually. Try it $\endgroup$
    – Dylan
    Apr 13, 2014 at 2:23

2 Answers 2

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Hint: The denominator is greater than $x^{1+2+\cdots+n}$, which is $x^{(n)(n+1)/2}$.

Thus the $n$-th term is $\lt \frac{1}{x^{n(n-1)/2}}$. Thus the $n$-th term is rapidly decaying. In particular, for $n\ge 3$, it is $\lt \frac{1}{x^n}$.

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Leaving out first $2$ terms doesn't affect the convergence, so $$\sum_{n=3}^{\infty} \frac{x^n}{(1+x)(1+x^2)...(1+x^n)}\le\sum_{n=3}^{\infty} \frac{x^n}{x\cdot x^{n-1}\cdot x^n}=\sum_{n=3}^{\infty} \left(\frac1x\right)^n$$ Which is a geometric series and converges because $\dfrac1x<1$.

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