/sci/ - Science & Math

[Anonymous No.6535974]

I hate asking /sci/ for hw help, but I don't have much experience with proving things.

I need help proving a summation inequality, I easily showed it was true for n=1, and assumed it was true for n=k, now the inductive step showing it is true for n=k+1, I have no clue how to go about it.

[Anonymous No.6535985]

1. Prove it's true for 1.
2. Prove it's true for 2.

Congratulations, you've proved it's true for n+1.

[Anonymous No.6535994]

>>6535985

fuck off

[Anonymous No.6536007]

>>6535974
well, add <div class="math"> \frac{1}{(n+1) ^2} </div> to both sides
then show that the right side of that inequality is less than or equal to the "n+1 version" of the right side

[Anonymous No.6536138]

The left side will never reach 2 it will keep getting closer and as it reaches closer to 2 the other side moves towards 2 so it will always be smaller

[Anonymous No.6536158]

>>6536138
silly rabbit
thats not how proof works

[Anonymous No.6536164]

show that it is true for n = 1 (trivial). Then, assume it is true for n=k. Adding 1/(k+1)^2 to both sides will give you the inequality you need, as 1/(k+1)^2 < 1/k.

[Anonymous No.6536226]

>>6535974


<span class="math"> \sum_{1}^{n+1}\frac{1}{k} =\sum_{1}^{n}\frac{1}{k}+\frac{1}{k+1}<2-\frac{1}{n}+\frac{1}{n+1} [/spoiler]

<span class="math"> =2-\frac{1}{n+1}\frac{(n+1)^2}{n^2}<2-\frac{1}{n+1} [/spoiler]

For n>0.
P(n)=>P(N=1)

[Anonymous No.6536230]

>>6536226
Fuck. I missed a bunch of squares out when i latexd it. But that's the gist

[Anonymous No.6536233]

>>6536226
Corrected

<span class="math">\sum_{1}^{n+1}\frac{1}{k^2} =\sum_{1}^{n}\frac{1}{k^2}+\frac{1}{(k+1)^2}<2-\frac{1}{n}+\frac{1}{(n+1)^2}[/spoiler]

<span class="math">=2-\frac{1}{n+1}\frac{(n+1)^2}{n^2}<2-\frac{1}{n+1}[/spoiler]