Math Test Grade: Did I get screwed?

[Woozie]

In what kind of situation would it matter where the constant started? I just checked a few calculus texts and they all seem to say the same thing me and oldoldman thought, though I can't seem to find a proof that the starting point is irrelevant.

Edit: And even if I am wrong, he should still get his points on the grounds that his calculus text probably says what mine says - that the start of the series doesn't matter. If the book printed wrong info, I don't think he should lose points for following what the book says (unless his teacher explicitly told them at some point that the book was wrong or explicitly told them that they have to use 1 on all of their tests).

I leave all of my more advanced calc books at work, but I'll check if I go to work today, or I'll ask one of the professors I work with. I'm pretty sure it doesn't matter, but maybe I'm wrong.

Edit 2: What the heck kind of professors do you guys have? Your professor took an entire 10% off because of a sign mistake? Was both of those situations you cited from the same professor? Maybe I'm just lucky enough to have forgiving professors.

[Woozie]

Looks like aalryn was right. I tried to write out a proof that it doesn't matter where the integral starts, but my proof only works when the integral starts in the same place as the sum's index. I can't seem to make it work for any arbitrary c (though I can't prove that starting at any arbitrary c is fallacious, and I can't come up with a counter example either. If someone knows one, please post it). So in your case, since the index started at 1, so should your integration.

Edit:

http://tutorial.math.lamar.edu/Class...egralTest.aspx

There are a couple of things to note about the integral test. First, the lower limit on the improper integral must be the same value that starts the series.

[oldoldman]

FYI That's just the way the image is notated on wikipedia, the way we notated replaced k=1 with n=1 and n with infinity. And the constant you start you integration as does not matter because while it may change the number you find the integral converging to it will not change whether the integral converges or diverges because the integral is going to infinity and ultimately if all you are checking is the end behavior you are just looking at what happens at infinity and what happens at C is irrelevant. (Provided there is no ln(0) or x/0 or anything like that happening at C, which I notated).

[Woozie]

That's the same thing I figured, but for some reason I can't prove it. The problem comes if one of the actual terms in the series is divergent. For example, if a3=5/0 or something like that, and you start the integral at 4, you could miss the divergence. I can't think of any continuous function that could work as a counterexample though. I can create strange functions that does (like: If x<3, f(x) =y, else f(x)=z or something like that). Of course, it doesn't seem like your problems (or most normal problems) are of that nature, so I still don't think you should have lost points for that. I'm not saying your wrong, I'm saying my previous statement was technically too broad (meaning I'm the one who's wrong). In your problems, it is clear that it wouldn't matter where the integration starts because your second problem diverged no matter what and your first is clearly convergent.

[oldoldman]

Update! I'm pretty sure I just aced the final which will give me an 89.4 for the class.

[Woozie]

Which means what your professor did was the difference between an A and a B? You really should point this out to him. Most professors I know would give you the difference, especially when the difference was due to something you shouldn't have lost points for in the first place.

[oldoldman]

Which means what your professor did was the difference between an A and a B? You really should point this out to him. Most professors I know would give you the difference, especially when the difference was due to something you shouldn't have lost points for in the first place.

The guy is ludicrously stubborn.

[Ramor]

Which means he's just as stubborn with everyone else so you shouldn't have anything to worry about

[Woozie]

Edit: Nevermind. I'm not making sense. I can't get my words out clearly. I noticed an error in one of my previous posts. I posted that there could be a divergence at or before C, but from the looks of it, you had already said that there were none in the post before mine (and you said you pointed that out on the test). I wasn't paying attention when I made my next post and just noticed it. So yeah, you were right.