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The only tests my class has learned so far and is allowed to use are: Divergence, Integral, Comparison, Geometric/Harmonic/Telescopic. I have proved that the series converges, via a ratio test, but my teacher doesn't want me to use that. I don't think I can do an integral test; it defnitely isn't a geometric series because the numerator increases arithmetically. It isn't any sort of p-series, not telescopic. That leaves the comparison test -- except I don't know what to compare it to. I tried $b_n = 1/3^n$, and while this converges, it's smaller than $a_n$.

As a side note, are there any tricks for finding good comparisons, or do you just develop an intuition over time?

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  • $\begingroup$ You can use the integral test, as you can integrate $f(x)=x3^{-x}$ (which is positive and decreasing for sufficiently large $x$) using integration by parts. $\endgroup$ – Mark McClure Apr 28 '16 at 1:22
  • $\begingroup$ We could compare with $1/(1.1)^n$. $\endgroup$ – André Nicolas Apr 28 '16 at 1:22
  • $\begingroup$ @AndréNicolas I don't understand. If I try to do a direct comparison, I can't conclude that $1/(1.1)^n$ is greater or smaller because the numerator is smaller ($b_n$ is smaller overall) but the denominator is also smaller ($b_n$ is greater overall). So I could use a comparison of $lim ( a_n / b_n)$, but I got stuck with $lim (1.1/3)^n * n$. $\endgroup$ – Alex G Apr 28 '16 at 1:36
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    $\begingroup$ We show that $\frac{n}{3^n}\lt \frac{1}{(1.1)^n}$ by showing $(3/1.1)^n\gt n$. Actually, we could use $1.5$, and show that $2^n\gt n$. $\endgroup$ – André Nicolas Apr 28 '16 at 1:49
  • $\begingroup$ Oh, I never even thought of manipulating it like that! My book typically just looks at the denominator or numerator and makes a decision based on that (eg. "denominator has less terms but same degree? it's smaller, so the number overall must be bigger" etc.). $\endgroup$ – Alex G Apr 28 '16 at 1:56
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You can combine techniques. For example $\sum b_n$ where $b_n=2^n/3^n$ converges as a geometric series. On the other hand, $n/3^n < b_n$, so your sum converges by the comparison test.

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  • $\begingroup$ Sorry, I can't seem to figure out how to prove that $2^n > n$. $\endgroup$ – Alex G Apr 28 '16 at 1:39
  • $\begingroup$ You could try induction. $\endgroup$ – Christian Gaetz Apr 28 '16 at 1:41
  • $\begingroup$ Working on it.... $\endgroup$ – Alex G Apr 28 '16 at 1:52
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    $\begingroup$ You are on the right track. Once you get to $2^{k+1}>2k$ just note that $2k \geq k+1$ since $k\geq 1$. Thus $2^{k+1}>2k \geq k+1$ which gives the induction step $\endgroup$ – Christian Gaetz Apr 28 '16 at 2:43
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    $\begingroup$ Thank you for actually putting in the work and showing your thoughts. Too many people on this site just want other people to do their homework for them. $\endgroup$ – Christian Gaetz Apr 28 '16 at 2:47

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