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With direct comparison you need both terms of both infinite sums to be greater than zero and one of the terms to be bigger or smaller. So graphing of those terms should work to see when they are greater than zero and when one term is bigger than another term. Problem I was told that I am not allowed prove things by comparing graphs and must actually prove that the graphs represent what is happening.

And I also do not understand why the derivative is taken to show when the first term is smaller(the ln(n)/x^2) is smaller than the second term(square root of x/x squared) all that shows when something is increasing or decreasing.

As shown in the graph below I think it should be x >= 1 and you can also do the inequality (Square root of x - lnx) > 0 and see when that is true meaning the first term is smaller than the second term when this is true but the only problem about this is that it does not tell you when it is positive. Graph representing each term at n in which n = x

So is it ok to use graphs like this to prove one term is smaller than the other term, and is the teachers logic taking the derivative and getting x>=4 correct?

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  • $\begingroup$ A graph is not a proof because how would you explain it to a blind person? Mathematics should not depend on if you have working eyes. Also how would you know if theres a bug in the graphing software? $\endgroup$ Commented Apr 29, 2023 at 1:48
  • $\begingroup$ You could manually plot it and tell the points (x,y) to the blind person the result would be the same. $\endgroup$
    – Golden Boy
    Commented Apr 29, 2023 at 1:59
  • $\begingroup$ You want to tell the blind person how many points? This would rely on the blind person trusting you, which is also not the point of a proof. $\endgroup$ Commented Apr 29, 2023 at 2:02
  • $\begingroup$ But in this case does the graph actually represent what is happening with the terms? I understand what you mean by the strict definition of proof for math. $\endgroup$
    – Golden Boy
    Commented Apr 29, 2023 at 2:09
  • $\begingroup$ In this case yes. But Desmos is not perfect; why are you trusting me when I say it is correct :) Here's an example where Desmos doesn't work right. And in any case, you interpreted the graph wrong. The maximal region of validity is in fact $x>0$. $\endgroup$ Commented Apr 29, 2023 at 2:12

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  1. A graph is not a proof because how would you explain it to a blind person? Mathematics should not depend on if you have working eyes.
  2. How would you know if theres a bug in the graphing software? Or if you are simply trying to plot something more complicated than Desmos can show? Here's an example where Desmos struggled. And here you can't really say that Desmos is wrong, but it doesn't show some important features of the function that OP was trying to plot.
  3. It is in fact correct that $\sqrt x > \ln x$ for all $x\ge 1$. This also means that it is true for all $x\ge 4$. Saying it is true for all $x\ge 4$ is not the same as saying it is not true for $x<4$. But you don't have a proof of the fact so you cannot use it.
  4. Actually, the inequality is true for all $x>0$, and this is in your graph, so I'm not sure why you chose $x\ge 1$.

but the only problem about this is that it does not tell you when it is positive. 5. Yes, that's correct. That's why you still need the next line. The derivative step shows that $\sqrt x - \ln x$ is increasing when $x>4$. So this proves that $\sqrt x - \ln x > \sqrt y - \ln y$ for all $x>y\ge 4$.

In particular this is true with $y=4$, and then you can check that $\sqrt 4 - \ln 4 > 0$. For instance, $\sqrt 4 - \ln 4 > 2 - \ln e = 1 > 0$ (because $\ln$ is increasing.) These two facts together give $\sqrt x - \ln x> 0$. And since we started this analysis with $x>4$, we cannot extend this to $x\le 4$ without further analysis. $x=4$ comes from the above direct computation, so all we can say we have proven is that $\sqrt x - \ln x > 0$ for $x\ge 4$.

  1. It isn't so hard to prove it for all $x>0$. You just need to use the second derivative test to show that $x=4$ is the unique local minimum.
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