NotNotBatman
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In my notes it says (*)if the limit of the partial sums  s_n is S, then the infinite series converges to S.; i.e  \sum_{k=p}^{\infty} a_k= S

Next it says (**) if the series converges to S, then  s_n \to S \quad \mathrm{and} \quad  s_{n-1} \to S \quad \mathrm{ as } \quad n\to \infty, \quad \mathrm{so} \quad a_n = s_{n} - s_{n-1} \to 0 \quad \mathrm{as} \quad n \to \infty

The converse is false.

But if the converse is false wouldn't that mean that the first statement (*) is false, because you would have the limit of partial sums as S, so the series should converge to S?

The harmonic series is used as an example with  1+\frac{1}{2} + (\frac{1}{2} + \frac{1}{4}) + \bigg{(} \frac{1}{5} +\frac{1}{6} +\frac{1}{7} +\frac{1}{9} \bigg{)} + \ldots \geq 1+\frac{1}{2} + \frac{1}{2} + \ldots , but I don't understand what this is implying.

Note, we are working in the complex plane.
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mqb2766
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The Harmonix series is probably the most famous example where the terms tend to zero, but their sum is infinite. I take it you understand the bracketing proof?
So what your notes are saying is that itiis necessary for the terms to converge to zero for the the sum to be finite, but this not sufficient, as the above example shows.
(Original post by NotNotBatman)
In my notes it says (*)if the limit of the partial sums  s_n is S, then the infinite series converges to S.; i.e  \sum_{k=p}^{\infty} a_k= S

Next it says (**) if the series converges to S, then  s_n \to S \quad \mathrm{and} \quad  s_{n-1} \to S \quad \mathrm{ as } \quad n\to \infty, \quad \mathrm{so} \quad a_n = s_{n} - s_{n-1} \to 0 \quad \mathrm{as} \quad n \to \infty

The converse is false.

But if the converse is false wouldn't that mean that the first statement (*) is false, because you would have the limit of partial sums as S, so the series should converge to S?

The harmonic series is used as an example with  1+\frac{1}{2} + (\frac{1}{2} + \frac{1}{4}) + \bigg{(} \frac{1}{5} +\frac{1}{6} +\frac{1}{7} +\frac{1}{9} \bigg{)} + \ldots \geq 1+\frac{1}{2} + \frac{1}{2} + \ldots , but I don't understand what this is implying.

Note, we are working in the complex plane.
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NotNotBatman
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(Original post by mqb2766)
The Harmonix series is probably the most famous example where the terms tend to zero, but their sum is infinite. I take it you understand the bracketing proof?
So what your notes are saying is that itiis necessary for the terms to converge to zero for the the sum to be finite, but this not sufficient, as the above example shows.
I get that all the brackets are bigger than 1/2 so it's bigger than a divergent series, so it's divergent.

However, i still dont understand, because if s_n -s_n-1 tends to 0, then they have the same limit, so lim(s_n) = sum of a_k by the first fact and the sum is finite, so it's sufficient?
(Apologies for formatting)
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mqb2766
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If the terms tend to zero and the sum is finite you're correct. However, the first condition is not enough by itself.
(Original post by NotNotBatman)
I get that all the brackets are bigger than 1/2 so it's bigger than a divergent series, so it's divergent.

However, i still dont understand, because if s_n -s_n-1 tends to 0, then they have the same limit, so lim(s_n) = sum of a_k by the first fact and the sum is finite, so it's sufficient?
(Apologies for formatting)
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NotNotBatman
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(Original post by mqb2766)
If the terms tend to zero and the sum is finite you're correct. However, the first condition is not enough by itself.
I'm reading that as "if the terms tend to 0 and the series is convergent, then the series is convergent", which is circular reasoning. I'm probably interpreting it incorrectly.
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B_9710
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(Original post by NotNotBatman)
I'm reading that as "if the terms tend to 0 and the series is convergent, then the series is convergent", which is circular reasoning. I'm probably interpreting it incorrectly.
You could have  s_n \rightarrow \infty and you could still have  s_n -s_{n-1} \rightarrow 0 .
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NotNotBatman
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(Original post by B_9710)
You could have  s_n \rightarrow \infty and you could still have  s_n -s_{n-1} \rightarrow 0 .
Right, I see what you mean.
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