0
$\begingroup$

It seems like so much of this is based on intuition and assumptions. I don't understand the limit comparison test contrapositives so I ignore the limit comparison test as it seems largely useless and cumbersome.

In trying to find the convergence or divergence of $\sum_1^\infty \frac{1}{\sqrt{n} + \ln n}$ why is it wrong to use the comparison test? I know that this can't converge because lnn is much smaller than $n^\frac{1}{2}$ so by p series it could never converge.

$\sum_1^\infty \frac{1}{\sqrt{n} + \ln n} > \sum_1^\infty \frac{1}{\sqrt{n} + n^\frac{1}{4}}$ and that $.5 + .25 < 1$ so by p series it diverges and it is a smaller function so by comparison test it diverges.

Everyone says this is wrong, but how accurate and precise do I need to be because it seems like a cumbersome approach like the limit comparison test makes just as many assumptions.

$\endgroup$
  • $\begingroup$ Why was the $n^\frac{1}{2}$ replaced by the $n^\frac{1}{4}$? $\endgroup$ – Makoto Kato Jul 30 '13 at 1:51
  • $\begingroup$ How does $.5 + .25 < 1$ show the series diverge? $\endgroup$ – Makoto Kato Jul 30 '13 at 1:59
  • $\begingroup$ @Makoto: As for the replacement, $\ln n$ is also smaller than $n^{1/4}$ for large enough $n,$ so that isn't particularly problematic. As for the inequality mentioned, it doesn't show that the series diverges, but his approach is still on the right track, even though it took a detour near the end. $\endgroup$ – Cameron Buie Jul 30 '13 at 5:46
  • $\begingroup$ @CameronBuie If he hadn't replaced $n^\frac{1}{2}$ by $n^\frac{1}{4}$, $\sum_{n=1}^\infty\frac1{n^{1/2}+n^{1/4}}$ would have been a lot simpler. $\endgroup$ – Makoto Kato Jul 30 '13 at 5:58
  • $\begingroup$ @Makoto: True. Still, he was nearly there, despite the slightly inefficient approach. $\endgroup$ – Cameron Buie Jul 30 '13 at 6:00
5
$\begingroup$

Unfortunately, you cannot apply the $p$-series result to $$\sum_{n=1}^\infty\frac1{n^{1/2}+n^{1/4}},$$ since it is not a $p$-series.

However, you're nearly there. Noting that for $n\ge 1$ we have $2n^{1/2}\ge n^{1/2}+n^{1/4}$ then we have $$\sum_{n=1}^\infty\frac1{n^{1/2}+n^{1/4}}\ge\frac12\sum_{n=1}^\infty\frac1{n^{1/2}}.$$ Now you can apply the $p$-series result.

$\endgroup$
  • $\begingroup$ But that isn't what I am saying. I am saying the opposite that it is impossible for those values to be larger than n. $\endgroup$ – Paul the Pirate Jul 30 '13 at 1:07
  • $\begingroup$ @PaulthePirate: You were quite close! See my updated answer. $\endgroup$ – Cameron Buie Jul 30 '13 at 1:40
  • $\begingroup$ Just to provide an outside opinion: in addition to being correct, this answer is quite close to what the OP was doing, hence shows that he was actually on the right track. (The bit about justifying divergence using $.5 + .25 < 1$ cannot be salvaged though, as far as I can see...) $\endgroup$ – Pete L. Clark Jul 30 '13 at 2:06
  • $\begingroup$ I don't understand why not. Isn't that always true? $\endgroup$ – Paul the Pirate Jul 30 '13 at 21:36
  • 1
    $\begingroup$ @Paul: It is certainly always true that $.5+.25<1,$ but I advise against focusing on that. What matters is that $\min\{.5,.25\}\le1.$ That's what allows us to pull the trick from my answer. Your condition certainly implies that $\min\{.5,.25\}\le1,$ but I fear that fact may mislead you, since (for a contrasting example) we have $5+.75> 1,$ and yet $$\sum_{n=1}^\infty\frac1{n^{1/2}+n^{3/4}}$$ diverges, using the same kind of reasoning as I gave above. $\endgroup$ – Cameron Buie Jul 30 '13 at 22:57

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.