/sci/ - Science & Math

>>14951090
Apologies i got confused again and thought there were only n terms. You're right, for a large value of [math] a [/math] the sum on the left can be bigger than the sum on the right, however you can find that each [math] x_n [/math] has a lower and upper bound using an inequality similar to the one in >>14950673
[eqn] \frac{an-1}{n^{2}+an}\le\frac{\left\lfloor an\right\rfloor }{n^{2}+\left\lfloor an\right\rfloor }\leq x_{n}\leq\frac{\left\lfloor an\right\rfloor }{n^{2}+1}\le\frac{an}{n^{2}+1} [/eqn]
And since [eqn] \lim_{n\rightarrow\infty}\frac{an-1}{n^{2}+an}=\lim_{n\rightarrow\infty}\frac{an}{n^{2}+1}=0 [/eqn]Then sequence of [math] x_n [/math] must converge to 0 by the squeeze theorem.

>>14951383
>>14951901
The rate of the reaction can be described as the change of the concentration of the reactant [A] with time[eqn] \frac{d[A]}{dt}=-k[A]^{n} [/eqn]Where [math]n[/math] is the order of the reaction. This is a separable differential equation which you can solve easily to end up with
[eqn] \begin{matrix} \frac{1}{[A]^{n-1}}=\frac{1}{[A]_{0}^{n-1}}+\frac{kt}{n-1} & \text{For $n\neq1$} \\ \ln[A]=\ln[A]_{0}-kt & \text{For $n=1$} \end{matrix} [/eqn]
Where [math] [A]_0 [/math] is the initial concentration. Therefore you should plot [math] [A], \ln[A], [A]^{-1}, [A]^{-2} [/math] against time, these plots correspond to zero, first, second and third order kinetics respectively. You should see which one of these plots gives a line to find the order of the reaction.