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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.