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yup 👍

yup basically just adding that limit notation to show that ur actually taking the infinite sum

if u do take the limit u get that the result would equal 1 + ln(2) + ln(2) > 1+ln2 u can take the limit from the start just by writing \lim_{k \rightarrow \infty} \sum_{n=2}^{k} \ln \left( \frac{n^2}{n^2-1} \right ) and if u want to expand the sum out to see the telescoping series the result...

infinity isnt a number most of the time when u see infinities in math ur actually just taking a limit to infinity. the infinity commonly used in math is just another way of saying what happens as this number gets really big. u can still make a telescoping sum if the symbol used is infinity it...

k is just an arbitrary value u can use any letter as long as u take the limit to infinity

yea my tutor gave me the textbook cause he thought i would have liked some of the questions. at first i tried using a trig substitution but i couldnt get anywhere with that so i was just wondering if there was another solution.

can someone check if my working is correct i got this integral from spivak's calculus \int \frac{x^2 -1}{x^2 +1} \frac{1}{\sqrt{1+x^4}} \; dx \quad \text{Using } u^2 = x^2 + \frac{1}{x^2} u^2 = x^2 + \frac{1}{x^2} \implies 2u \; du = \left(2x - \frac{2}{x^3}\right)dx u \; du = \left(x -...

nah ive seen an algebraic proof of it in a paper before and got stuck on the reverse triangle inequality

probably nice to know the proof of the triangle inequality and the reverse triangle inequality

derivatives of inverse functions are in the yr 12 mx1 scope not the yr 11 one if i recall correctly

p does not equal mv anymore :(

i just realised this is invalid since u cant apply the binomial theorem once again... well another approach u could do is RHS-LHS RHS-LHS = \frac{3^k}{2^{2k}} - \frac{1}{2^{k+1}} - \frac{1}{2} =\frac{3^k - 2^{k-1} - 2^{2k-1}}{2^{2k}} =\frac{3^k -2^{k-1}\left(1+2^k\right)}{2^{2k}} Note that 3^k...

also i dont think u would be able to apply induction here since 0<k<1 and usually u can only apply induction when k is an integer

if you wanted to do the question without induction btw heres the working out: \frac{1}{2}+\frac{1}{2^{k+1}}=\frac{1}{2}\left(1+\frac{1}{2^k}\right) =\frac{1}{2}\left(\frac{2^k + 1}{2^k}\right) =\frac{2^k + 1}{2^{k}(2)} Notice that 3^k>2^k +1 by the binomial theorem as: 3^k =(2+1)^k = 2^k...

actually nvm i dont think this proof is valid as well since 0<k<1 oops.

for the specific case of n=2 however u could probably just use the binomial theorem writing 3/4 as 1/2 +1/(2^2) (\frac{3}{4})^{k}=(\frac{1}{2}+\frac{1}{2^2})^k =\frac{1}{2^k} + \binom{k}{1}\frac{1}{2^{k+1}} + \binom{k}{2}\frac{1}{2^{k+2}} + ... + \frac{1}{2^{2k}} > \frac{1}{2^k} +...

is there any information on what n is? i dont think this would be a true statement if theres no restriction on n which can be seen in this desmos graph

Another way to do it algebraically is to represent the complex numbers as cis(a1) and cis(a2) then use the sum to product trigonometric identities.

yea its all goods imo the whole point of this forum is just to communicate ideas about math and how they can help other people so its not really unexpected u get informal proofs and algebruh errors

the conjugate of \frac{1}{a+bi} is not \frac{a-ib}{a^2+b^2} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2} so the conjugate is \frac{a+bi}{a^2+b^2} so using this method u still get the same answer as i did as u get \frac{(x-iy)(a+bi)}{a^2 + b^2} =\frac{(x-iy)(a+bi)}{(a+bi)(a-bi)} =\frac{x-iy}{a-bi}