Sum of bounds of \sum \frac{1}{a^{n}+b^{n}}

tripiverded9

tripiverded9

Answered question

2022-01-13

Sum of bounds of 1an+bn

Answer & Explanation

ol3i4c5s4hr

ol3i4c5s4hr

Beginner2022-01-14Added 48 answers

We can suppose b<a. ll assume too that the sum begins in n=1, so the obvious lower bound (with b=0) is n=11an=1a1. I don´t know if there is an exact way to find the sum, but one way I found to get upper/lower bounds as tight as you want uses this equality:
1an+bn=1anbnan1(an+bn).
Using it, you can deduce:
1an+bn=1anbnan1(an+bn)=1anbna2n+b2na2n1an+bn=
=1anbna2n+b2na3nb3na3n1an+bn=
Calling sk=bknakn1(an+bn), it´s easy to see that sk tends to 0 when k goes to infinity (as the greater sequence
tk=bknakn goes to 0).
So, 1an+bn=1anbna2n+b2na3n+=
Tiefdruckot

Tiefdruckot

Beginner2022-01-15Added 46 answers

We do have this, in terms of Jacobi theta functions:
For a>1>b>0,
n=+1an+bn
v2(ilog(a)2,bav2(0,ba)v3(0,ba)b4(0,ba)
2iv1(ilog(a)2,bav1(π2,ba)

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