# Sketchy Induction Proof

• Sisyphus

#### Sisyphus

something funny's going on here, and I can't see what :yuck:

For a sequence $${x_n}$$ , where each term is non-negative

the series $$x_1 + x_2 + ... +x_n + ...$$ converges

proof:

it will suffice to show that the sequence of partial sums $${s_n}$$ is bounded

where each $$s_i = x_1 + ... + x_i$$

when i=1,
$$s_1 = x_1$$

so the result holds true for i=1

let the result be true for all positive numbers up to some k such that

$$s_k <= some b$$

now consider $$s_{k+1}$$...

$$s_{k+1} = s_k + x_{k+1} <= b + x_{k+1}$$

so the result holds true for all k= 1, 2, 3 ...

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Of course finite sums are always bounded!

You're implicitly implying that there are infinitely many partial sums. Induction doesn't work for infinity, it only works for a given natural number.

ie, you haven't shown that the supremum of all partial sums exists.

it will suffice to show that the sequence of partial sums $${s_n}$$ is bounded
no it won't

You need to show: there exist a S such that for any given Epsilon greater than 0 there exist a N (element of the natural numbers) where |S_n - S| < Epsilon provided that n>N.

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no it won't

You need to show: there exist a S such that for any given Epsilon greater than 0 there exist a N (element of the natural numbers) where |S_n - S| < Epsilon provided that n>N.

Well, since the terms in the partial sums are non-negative all he has to show is that ALL the partial sums are bounded by some number. And then he has a bounded monotone sequence of partial sums which implies they converge.

The induction showed that for every partial sum s_n there existed a number b such that s_n<b. This DOES NOT mean there exists an m such that s_n<=M for all n.

ah, thanks for clearly that up. That's kind of subtle, and I doubt I would've figured it out by myself..

i guess i forgot that one of the reasons why we speak of partial sums in the first place is that they are all bound.. Induction doesn't work that way! Induction on n shows that statement Sn is true for any finite n.