Yiuyu
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Prove from the definition of the phrase " a real series converges to a real number L" how the sum of n=1 to infinity (1/10^n) converges to 1/9.
How would I do this using the defn?
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Prasiortle
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(Original post by Yiuyu)
Prove from the definition of the phrase " a real series converges to a real number L" how the sum of n=1 to infinity (1/10^n) converges to 1/9.
How would I do this using the defn?
Firstly, do you know and understand what the definition is saying?
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Yiuyu
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(Original post by Prasiortle)
Firstly, do you know and understand what the definition is saying?
Yes?
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RDKGames
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(Original post by Yiuyu)
Yes?
If you consider the sequence of partial sums u_n = \{ S_1, S_2, S_3, \ldots, S_n \} where S_n = \displaystyle \sum_{r=1}^{n} \dfrac{1}{10^r} then the limit (u_n) \xrightarrow{n \to \infty} L will be the answer to the actual sum S_{\infty}.
So you just want to show that the sequence of partial sums has a limit of L=\dfrac{1}{9}
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DFranklin
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(Original post by Prasiortle)
Firstly, do you know and understand what the definition is saying?
Despite what he says, my feeling is "no, not really". He previously posted asking for the definition and saying he'd tried using google, despite the fact that googling gives lots of hits with no difficulty at all.

So my guess would be "technically, he knows what the definition says, but he doesn't know how to show a sequence converges, so... :shrugs:". [I'm afraid that after helping on here for 10 years, you get cynical].

(Original post by NotNotBatman)
I'm not entirely sure what the definition is saying, just that the real series converges ?

But  \frac{1}{10^n} = (\frac{1}{10})^n and \sum_{n=1}^{\infty}x^n converges for |x|<1

so you know it converges, so you can use the sum to infinity formula.
No, you can't. With the greatest respect, if you don't understand what the definition is saying, and the question says "prove from the defintion...", it is perhaps not the most sensible thread for you to comment on.
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NotNotBatman
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(Original post by DFranklin)
Despite what he says, my feeling is "no, not really". He previously posted asking for the definition and saying he'd tried using google, despite the fact that googling gives lots of hits with no difficulty at all.

So my guess would be "technically, he knows what the definition says, but he doesn't know how to show a sequence converges, so... :shrugs:". [I'm afraid that after helping on here for 10 years, you get cynical].

No, you can't. With the greatest respect, if you don't understand what the definition is saying, and the question says "prove from the defintion...", it is perhaps not the most sensible thread for you to comment on.
What I don't understand is how that is a definition at all.
" a real series converges to a real number L" to me is a statement that just means that there is a real series that converges.

If that's wrong, what does it mean?
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DFranklin
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(Original post by NotNotBatman)
What I don't understand is how that is a definition at all.
" a real series converges to a real number L" to me is a statement that just means that there is a real series that converges.

If that's wrong, what does it mean?
Ok, to be fair, this is something that's hard to know without university context.

If you google "series convergence definition" (not in quotes), you will find that there is a precise definition (well, there are various definitions, but they all mean the same thing) of what it actually means when you say "a real series converges to a real number L".

When the question says "from the definition of the phrase...", it means that you need to prove the result using that definition (and specifically, NOT using results that are proved after the definition).
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NotNotBatman
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(Original post by DFranklin)
Ok, to be fair, this is something that's hard to know without university context.

If you google "series convergence definition" (not in quotes), you will find that there is a precise definition (well, there are various definitions, but they all mean the same thing) of what it actually means when you say "a real series converges to a real number L".

When the question says "from the definition of the phrase...", it means that you need to prove the result using that definition (and specifically, NOT using results that are proved after the definition).
Ah, when it said "from the definition...", I was thinking epsilon - delta, but that wouldn't make sense obviously. But now my memory of partial sums is rejogged. I'll edit my post.
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DFranklin
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(Original post by NotNotBatman)
Ah, when it said "from the definition...", I was thinking epsilon - delta, but that wouldn't make sense obviously. But now my memory of partial sums is rejogged. I'll edit my post.
Well, they actually do mean epsilon-delta I'm sure.

It's a bit ambiguous, because you can either say "a series converges if the sequence of partial sums converge" (in which case the epsilon-delta's don't come in until you worry about proving the sequence converges), or you can directly plug in the partial sums into the sequence definition (so you get a definition looking like: \forall \epsilon > 0, \, \exists N \text{ s.t. } n > N \implies |\sum_1^n a_n - S| < \epsilon).

If you don't go directly to epsilon-deltas, arguably you might have a little leeway in what you can claim: you could probably get away with using "a bounded monotonic sequence converges" without proving it.

It depends a little on the exact route your course takes with sequence / series convergence.
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