Solving a real analysis problem from a first year ACTUAL exam paper from the University of Cambridge. 🎓 www.jpimathstutoring.com 📷 / jpimaths ⏰ / @jpimaths Contact me: jpimaths@gmail.com
Cambridge math graduate here. This is a short question (seen by section I), meaning it should ideally take 5-15 minutes and is worth a total of 10 marks (however the mark distribution is not explicit). U also get long questions which are worth 20 marks and have a significantly greater impact on your final grade (the whole grading system is a bit too long and complicated for me to explain fully). There are 8 courses in first year, examined over 4 papers with 2 courses of questions in each. For example, paper 1 only has questions on Analysis and Vectors & Matrices. Each paper has 4 short questions and 8 long questions (2,4 for each course), where u can answer all short questions and at most 5 long questions in the 3 hours u get. Yes it is a very stressful experience 😭 Getting into second year is easy, all u need to do is not fail (only like 1 or 2 people fail out of the 250, but a few more tend to drop out).
Nice! Slightly differently, a) Since Σ(a^2) < (Σ(a))^2, so Σ(a^2) converges. b) Σ(a/2+b/2) converges and a/2+b/2 >= sqrt(ab). Therefore, Σ(sqrt(ab)) converges.
@@johnlv12 it's really cool right! It's a very weird feeling walking through the beautiful city filled with old buildings in sub fusc. Makes me feel as if I'm in Harry Potter!
@@JPiMaths i think it was something like given that sum from 1 to infinity of a_n converges, prove that sum from 1 to infinity of (a_n)^p converges for all p >= 1. I basically had the same proof as you
hello, im looking to apply to get in to oxford like you did, could you make some videos answering common maths oxford interview questions and MAT questions please? it would help a lot :)
@@user-mu6yz2db9w amazing! Best of luck with the application. Will definitely make lots of videos on this. I've already made a fair few: ru-vid.com/group/PL-yrvlQ9Z8t-zQOUDQPdyiKv2TMA0Q4V7
@@goyaldev09 great question! We know from the first part of c that if we increase p even by the slightest, the series will converge. Using this as motivation, we need a(n) to be, something similar to 1/n. Now, this of course diverges, so we need to slow down it's growth in order to ensure that sum(a(n)) converges. However, we know the harmonic series diverges very slowly so we won't need to slow it down too much more in order for it to converge. What terms can reduce the size of something, but only just?... 1/(log(n)). That kind of explains the thought process. However, in actuality I imagine a lot of my thinking was in my subconscious having seen lots of these sorts of things before and just having a good gut feeling!
maybe this is a stupid question by why can you make the claim that a_n < 1 given n > N in the first part? Since we only know that its positive and converges we cant say it converges to 0 or a number below 1, so isnt the only claim we can make is a_n < 1 + a, where lim(a_n) = a?
@@quite6461 so the definition of a sum converging is that its partial sums converge. Ie, if we define a_1+a_2+...+a_n to be s_n then sum(a_n) converges if and only if the limit as n goes to infinity of s_n exists. In order for such a limit to exist we need a_n to tend to 0 as n tends to infinity (or otherwise s_(n+1)-s_n wouldn't tend to 0). The sun of the a_n could very well converge to a number greater than 1, but purely from the fact that the sum converges, we know eventually the terms must be very small
You'll laugh at this: I proved (c) before watching the video. I did it from first principles, using the comparison test. The operation of taking aₙ to n⁻ᵖ√aₙ actually converts the harmonic series into a convergent one as long as p>1/2. That lets you compare aₙ to that "fixed" series, as it has to decay to 0 at the very least slower than the harmonic series does. But I didn't even think to use part (b) of this problem to prove part (c). Talk about using a machine gun when I could've used a fly swatter that was handed to me lol
