A Friend in Need.... 1) Your friend is saying something VERY wrong here. While it's true that the integral converges, that is NOT the reason. Certainly, functions that DONT go to zero cannot have convergent improper integrals, but improper integrals of functions that DO go to zero can do anything they want (just like a limit of type 0 · 1). This is the whole reason that we study improper integrals and series!!! You should explain to your friend that it's true that any positive decreasing function like this one will have an increasing concave down antiderivative. Yet, some increasing concave down functions go to infinity e.g. ln(x) while some never even reach one e.g. 1 1− x . The limit she evaluated only tells you that the slope of the antiderivative is even- tually trending toward zero- the question is: is it doing so quickly as the antiderivative approaches a finite vertical asymptote, or is it doing so slowly as the antiderivative increases to infinity? To do the problem correctly, your friend should integrate the function and take the limit. Alternatively, she could argue that since this is a continuous decreasing function, P1 1 n and n=1( e ) is a convergent geometric series, this integral must converge by the integral test. R 1 dx 2) 0 x2 converges by the p-test. R 1 dx Your friend did not notice that the integral was starting at zero. 0 xp does not 1 converge for any p. While x2 is good at getting to zero as x approaches infinity, it is symmetrically good at rushing up to infinity as x approaches zero. R 1 dx R 1 dx R 1 dx Since 0 x2 = 0 x2 + 1 x2 = divergent + convergent = divergent P P P 3) The rule that an +bn = an + bn only applies when all of the sums involved con- verge. This was a homework exercise you may recall. She should use partial fractions 3 to make this into a telescoping sum that goes to 4 cos(πn) 4) These terms are not alternating! In order to use the alternating series test, n would have to be positive for all n. In reality, cos(πn) = (−1)n, so every term of 1 P1 (−1)ncos(πn) P1 1 this series actually equals n . Thus, n=1 n = n=1 n which is a divergent p-series. 5) While it's true that x = 1 makes this a geometric series, any value of x would make P1 xn P1 x n this geometric! Which values of x will converge? Well, n=1 2n = n=1( 2 ) , which x converges if and only if j 2 j < 1 i.e. −2 < x < 2 6) Oh no, don't break up the denominator! Honestly, graders hate that. Doing so can easily make a problem impossible to solve (especially when you're doing integrals), or so easy that the solution is trivial. This series could easily be directly or limit compared P1 1 with n=1 n3 , a convergent p-series. 7) This formula only applies when jrj < 1 and the series converges! This series is very divergent by the nth term test. Furthermore, her indexes and powers don't match up for the geometric series formula. 8) Hm, looks like she mixed up the statement of the Ratio Test. You are supposed to look at the limit of an+1 , but she is looking at the reciprocal. After flipping her fractions an around, the limit will be 0 and the series converges by the ratio test. She should have realized that n! is MUCH larger than any power of n in the long term, so the series really ought to converge. 9) Oh noooooo! Every positive series has an increasing partial sum, and they certainly don't all diverge! She should know that not every increasing sequence diverges (in particular, we have a theorem that any bounded monotonic sequence converges). Sim- ilarly, not every increasing function goes to infinity. It matters a LOT how quickly the terms of the sequence of partial sums are increasing. You should show her that P1 1 n=1 n2 has increasing partial sums but the series converges, so she'll see the flaw in her reasoning. She should have said the sum diverged since it was a p-series with p ≤ 1 10) Her inequality is false. When you make the denominator bigger, the fraction gets smaller. She would be better off avoiding inequalities entirely by using the limit com- parison test. The worst mistakes seem to be in questions 1 and 9 they show that your friend really does not understand how series behave (or how functions can behave as the go toward infinity). Thinking that 7 converges and 8 diverges is not a great sign either. You should always take a moment to consider what you know about the relative strengths of the numerator and denominator before you begin a problem. Usually, reasoning to yourself why the series con- verges will actually give you a hint as to which test to use! (especially if a limit comparison is possible) Your friend should definitely talk to her instructor. She doesn't seem to really under- stand why some series converge and others diverge, or how and why the convergence tests work. With her instructor, she should make a plan to catch up that includes a one-on-one explanation, extra exercises, reading the book, office hours, etc. She should also try going to the math support center and attending Math 1102 for the rest of the semester..