I am trying to prove that the series &#x2211;<!-- ∑ --> <mstyle displaystyle="true" scriptlevel

cazinskup3

cazinskup3

Answered question

2022-06-22

I am trying to prove that the series 1 ( m 1 2 + m 2 2 + + m r 2 ) μ in which the summation extends over all positive and negative integral values and zero values of m 1 , m 2 , , m r , except the set of simultaneous zero values, is absolutely convergent if μ > r 2

Answer & Explanation

arhaitategr

arhaitategr

Beginner2022-06-23Added 13 answers

Here is a way I like. We can rewrite your sum as
m Z m { 0 } 1 m 2 r + ϵ
where ϵ > 0. Then since
x 2 max i | x i | ,
by using the comparison test, we know that our original series will converge if
m Z m { 0 } 1 max i | m i | r + ϵ
converges. Since the set of all m with k 1 max i | m i | k has size C k r 1 for some constant C (it is the surface of an r dimensional cube) we see that the above is bounded by
k = 1 C k r 1 k r + ϵ C k = 1 1 k 1 + ϵ
which converges.

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