MHB Converging Geometric Series with Negative Values?

Lancelot1
Messages
26
Reaction score
0
Hiya everyone,

Alright ?

I have a simple theoretical question. In a decreasing geometric series, is it true to say that the ratio q has to be 0<q<1, assuming that all members of the series are positive ? What if they weren't all positive ?

Thank you in advance !
 
Mathematics news on Phys.org
A "geometric series" is of the form \sum ar^n= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot= a(1+ r+ r^2+ r^3+ \cdot\cdot\cdot) so, yes, if the series is decreasing and positive then r must be less than 1. If r> 1 then 1&lt; r&lt; r^2&lt; r^3&lt; \cdot\cdot\cdot. If r< 1 then 1&gt; r&gt; r^2&gt; r^3&gt; \cdot\cdot\cdot. Of course, in the first case, r> 1, the series does not converge.

If they are not all positive then either they are all negative and we can take the negative into the "a" term so that \sum at^n= a\sum r^n has a negative number times the same sum of ar^n or they are alternating, \sum a(-r)^n= a\sum (-1)^n r^n. The geometric series \sum ar^n converges if and only if -1&lt; r&lt; 1.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Replies
3
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
6
Views
3K
Replies
14
Views
2K
Back
Top