# Convergence of $q_{1}+q_{1}^{2}+q_{2}^{3}+q_{1}^{4}+q_{2}^{5}+q_{3}^{6}+q_{1}^{7}+\ldots$

I am trying to show that the series $$q_{1}+q_{1}^{2}+q_{2}^{3}+q_{1}^{4}+q_{2}^{5}+q_{3}^{6}+q_{1}^{7}+\ldots$$ where $q_{n}=q^{1+\left( \dfrac {4} {n}\right)}$, and $\left(0<q<1\right)$ is convergent, although the ratio of the $(n+1)$th term to the $n$th term is greater that unity, when n is not a triangular number. I am actually having a hard time recognizing the pattern in the series itself hence am struggling with, what the $(n+1)$th term to the $n$th really look like.

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Note that $q_n \leq q$ for all $n$, since $q < 1$. – user17762 Mar 11 '12 at 21:54
@SivaramAmbikasaran Interesting observation but should n't the inequality be the other way around in your comment. – Hardy Mar 11 '12 at 21:58
since $q<1$, $q^{4/n} < 1$ Hence, $q^{1+4/n} < q$. – user17762 Mar 11 '12 at 22:02
I could be wrong here, but should n't that only hold for 4 > n ? as beyond that we 'd be taking roots of q in which case – Hardy Mar 11 '12 at 22:05
For instance, if $q<1$, we still have $\sqrt{q}<1$. – user17762 Mar 11 '12 at 22:06

As Sivaram noted, $q_n\le q$ for all $n$. Thus the series converges because it is dominated by the geometric series for $q$, which converges.