At what point do the sums of the reciprocals converge?by thegreatjared Tags: convergence, divergence, harmonic series, reciprocals, summations 

#1
Aug1711, 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?




#2
Aug1711, 09:16 PM

P: 2,070

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. 



#3
Aug1711, 09:18 PM

P: 5

Wow, simple as that. Thanks!!




#4
Aug1711, 09:48 PM

Mentor
P: 16,613

At what point do the sums of the reciprocals converge?
Also this should be interesting: http://en.wikipedia.org/wiki/Small_s...mbinatorics%29



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