Induction: How to prove propositions with universal quantifiers? In my book, they prove with mathematical induction propositions with successions like this:
$$1 + 3 + 5 + \cdots + (2n-1) = n^2$$
In all exercises. However, recently I took some exercises from a different paper and instead of these it told me to prove this:
$$\forall n \in N (11 / (10^{2n+1} + 1 ))$$
Or perhaps this:
$$ \forall n \in N (n < 2^n) $$
(I can't find how to write the natural numbers set symbol. $N$ is it there.)
And now I'm lost. This is is what I did with the first one:
Prove that the proposition works for $n=1$
$$ 11 / (10^3+1) \implies 11/1001
\implies \exists x \in N(11x=1001)$$
Which is true, if you take $x = 91$.
Assume
$$\forall n \in N (11 / (10^{2n+1} + 1 ))$$
We have to prove:
$$\forall n \in N (11 / (10^{2n+3} + 1 ))$$
We prove it:
Which I don't know how to do. Curiously enough, my book only shows exercises with successions, so I guess that this exercise can be, somehow, written as a succession? I am not sure about that. Any ideas?

I'm not sure what you mean by "writing a proof as a succession". I can do it as follows:
$$10^{2n+3}+1=100\cdot 10^{2n+1}+1=99\cdot 10^{2n+1}+\left(10^{2n+1}+1\right)$$
Look at the rightmost expressions: obviously, the first summand is a multiple of $\,11\,$ since $\,99\,$ is, and the second one is also a multiple of $\,11\,$ by the inductive hypothesis. Since the sum of two multiples is again a multiple we're done.
