Best solutions to - "Sum of bounds of ∑1an+bn"

tripiverded9

Answered question

2022-01-13

Sum of bounds of

\sum \frac{1}{a^{n} + b^{n}}

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

\sum_{n = 1}^{\infty} \frac{1}{a^{n}} = \frac{1}{a - 1}

. 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:

\frac{1}{a^{n} + b^{n}} = \frac{1}{a^{n}} - \frac{b^{n}}{a^{n}} \frac{1}{(a^{n} + b^{n})}

.
Using it, you can deduce:

\sum \frac{1}{a^{n} + b^{n}} = \sum \frac{1}{a^{n}} - \sum \frac{b^{n}}{a^{n}} \frac{1}{(a^{n} + b^{n})} = \sum \frac{1}{a^{n}} - \sum \frac{b^{n}}{a^{2 n}} + \sum \frac{b^{2 n}}{a^{2 n}} \frac{1}{a^{n} + b^{n}} =

= \sum \frac{1}{a^{n}} - \sum \frac{b^{n}}{a^{2 n}} + \sum \frac{b^{2 n}}{a^{3 n}} - \sum \frac{b^{3 n}}{a^{3 n}} \frac{1}{a^{n} + b^{n}} =

Calling

s_{k} = \sum \frac{b^{k n}}{a^{k n}} \frac{1}{(a^{n} + b^{n})}

, it´s easy to see that sk tends to 0 when k goes to infinity (as the greater sequence

t_{k} = \sum \frac{b^{k n}}{a^{k n}}

goes to 0).
So,

\sum \frac{1}{a^{n} + b^{n}} = \sum \frac{1}{a^{n}} - \sum \frac{b^{n}}{a^{2 n}} + \sum \frac{b^{2 n}}{a^{3 n}} + \dots =

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Tiefdruckot

Beginner2022-01-15Added 46 answers

We do have this, in terms of Jacobi theta functions:
For

a > 1 > b > 0

\sum_{n = - \infty}^{+ \infty} \frac{1}{a^{n} + b^{n}}

v_{2} (i \frac{\log (a)}{2}, \sqrt{\frac{b}{a}} v_{2} (0, \sqrt{\frac{b}{a}}) v_{3} (0, \sqrt{\frac{b}{a}}) b_{4} (0, \sqrt{\frac{b}{a}})

2 i v_{1} (i \frac{\log (a)}{2}, \sqrt{\frac{b}{a}} v_{1} (\frac{π}{2}, \sqrt{\frac{b}{a}})

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