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OT: golden ratio

his ratio sequence. Even he should

the sequence turn out to be the way we

ng

vergence

Haven't looked at the wiki proofs but there seems to be some kind of confus ion over series convergence. Formally, the series comprised of terms s1, s2 , s3,... is said to converge to a limit L if the series of partial sums S1, S2, S3, converges to L in the conventional analytic sense, where the parti al sum Sk=s1+s2+...+sk.

The harmonic series is trivial to analyze when you can do things like take sums over say 1/(N+1) + 1/(N+2) + 1/(N+N) > 1/(N+N) + 1/(N+N) + ...+1/(N+N) = N x 1/(N+N)= 0.5. Since there are an infinite number of non-overlappi ng subseries of this type, there are an infinite number of sums exceeding 0 .5, which, being a lower bound for the full series sum, sounds like an unbo unded sum to me.

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bloggs.fredbloggs.fred
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Sadly not :(

It diverges because we can put a trivial lower bound on the partial sum over any decade being definitely > 0.9 in the limit N -> infinity.

For the sub range N+1 to 10N the smallest term will be the last one

phi10N = 1/10N (1-1/((10N)^2+2))

There are 9N terms from N+1 to 10N so the sum must be at least

0.9(1 - 1/(100N^2+2))

Multiply up by 9N terms as a crude lower bound and it is obvious that the series will diverge since every decade contributes *at least* +0.9 to the sum and there are a heck of a lot of them out to infinity!

Unfortunately you seem unable to grasp the concept.

--
Regards, 
Martin Brown
Reply to
Martin Brown

Hi,

I understand the function diverges, however I think it makes sense to have categories of divergent functions based on the relative rates they diverge. For example for this function the rate the series has to grow to maintain a steadily decreasing divergence is exponential. That is a lot different than a divergent series that could grow linearly and have an exponentially growing divergence etc. Just saying a series is divergent or convergent misses out on this detail.

cheers, Jamie

Reply to
Jamie M

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