Tagiuraoob

## Answered question

2022-02-24

Evaluate
$\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{7}^{n}+1}$

### Answer & Explanation

Jonas Burt

Beginner2022-02-25Added 4 answers

7 is not special at all. We can replace it with any $a>1$ (even real numbers).
$\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{a}^{n}+1}=\sum _{n=1}^{\mathrm{\infty }}\varphi \left(n\right)\frac{{a}^{-n}}{1+{a}^{-n}}=\sum _{n=1}^{\mathrm{\infty }}\varphi \left(n\right){a}^{-n}\left(1-{a}^{-n}+{a}^{-2n}-\dots \right)$
$=\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^{\mathrm{\infty }}\varphi \left(n\right){\left(-1\right)}^{k+1}{a}^{-kn}$
$=\sum _{m=1}^{\mathrm{\infty }}\sum _{d\mid m}{\left(-1\right)}^{\frac{m}{d}+1}\varphi \left(d\right){a}^{-m}$
$=-\sum _{m=1}^{\mathrm{\infty }}{a}^{-m}\left(\sum _{d\mid m}{\left(-1\right)}^{\frac{m}{d}}\varphi \left(d\right)\right)$
Now we will show the following:

If m is odd, then $\sum _{d\mid m}{\left(-1\right)}^{\frac{m}{d}}\varphi \left(d\right)+\sum _{d\mid t}\varphi \left(2d\right)+\dots +\sum _{d\mid t}pfi\left(d\right)=-m$. If m is even, let $m={2}^{k}t$ for odd t and $k\ge 1$. Then
$\sum _{d\mid m}{\left(-1\right)}^{\frac{m}{d}}\varphi \left(d\right)=\sum _{d\mid t}\varphi \left(d\right)+\sum _{d\mid t}\varphi \left(2d\right)+\dots +\sum _{d\mid t}\varphi \left({2}^{k-1}d\right)-\sum _{d\mid t}\varphi \left({2}^{k}d\right)$
$=\left(1+\varphi \left(2\right)+\dots +\varphi \left({2}^{k-1}\right)-\varphi \left({2}^{k}\right)\right)\sum _{d\mid t}\varphi \left(d\right)=0$
Hence the sum equals to

Tail3vn

Beginner2022-02-26Added 3 answers

Another approach is to use Lambert Series generating function:
$\frac{q}{{\left(1-q\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right){q}^{n}}{1-{q}^{n}}=\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{q}^{-n}-1}|q\mid <1⇒$
$\frac{{q}^{2}}{{\left(1-{q}^{2}\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{q}^{-2n}-1}=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\left[\frac{\varphi \left(n\right)}{{q}^{-n}-1}-\frac{\varphi \left(n\right)}{{q}^{-n}+1}\right]$
Hence:
$\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{q}^{-n}+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{q}^{-n}-1}-2\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{q}^{-2n}-1}$
$=\frac{q}{{\left(1-q\right)}^{2}}-\frac{2{q}^{2}}{{\left(1-{q}^{2}\right)}^{2}}=q\frac{1+{q}^{2}}{{\left(1-{q}^{2}\right)}^{2}}$

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