Register to reply

At what point do the sums of the reciprocals converge?

Share this thread:
thegreatjared
#1
Aug17-11, 09:06 PM
P: 5
It is well known that the Harmonic Series diverges (1/1+1/2+1/3+1/4+...), but that the series [1/1+1/4+1/9+1/16+...] converges. In the first series, the denominators are the integers, whereas in the second example, the denominators are the integers to the power of 2. My question is, at what power do the sums of the reciprocals "switch" from divergence to convergence?
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
phyzguy
#2
Aug17-11, 09:16 PM
P: 2,179
The integral test for convergence tells us that the infinite sum:
[tex]\sum_1^\infty f(n)[/tex]
and the integral:
[tex]\int_1^\infty f(n)[/tex]
either both converge or both diverge. Since:
[tex]\int_1^\infty \frac{1}{x^{1+\delta}} = \frac{1}{\delta}[/tex]
This tells us that the infinite series:
[tex]\sum_1^\infty \frac{1}{x^{1+\delta}}[/tex]
will converge as long as delta is greater than zero, no matter how small it is.
thegreatjared
#3
Aug17-11, 09:18 PM
P: 5
Wow, simple as that. Thanks!!

micromass
#4
Aug17-11, 09:48 PM
Mentor
micromass's Avatar
P: 18,299
At what point do the sums of the reciprocals converge?

Also this should be interesting: http://en.wikipedia.org/wiki/Small_s...mbinatorics%29


Register to reply

Related Discussions
(revised+re-post)Upper and Lower sums & Riemann sums Calculus & Beyond Homework 3
Sequences, Cumulative Sums, Partial Sums Plot - Ti-89 Titanium Calculators 0
Sums of Reciprocals of Infinite Subsets of Primes General Math 16
Equation of a relation that appears to be its own reciprocal? Linear & Abstract Algebra 3