Evaluate∑n=1∞ϕ(n)7n+1

Question

Jonas Burt · Accepted Answer

7 is not special at all. We can replace it with any a&amp;gt;1 (even real numbers).  ∑n=1∞ϕ(n)an+1=∑n=1∞ϕ(n)a−n1+a−n=∑n=1∞ϕ(n)a−n(1−a−n+a−2n−…)  =∑n=1∞∑k=1∞ϕ(n)(−1)k+1a−kn  =∑m=1∞∑d∣m(−1)md+1ϕ(d)a−m  =−∑m=1∞a−m(∑d∣m(−1)mdϕ(d))  Now we will show the following:  ∑d∣m(−1)mdϕ(d)={0m  is even−mm  is odd    If m is odd, then ∑d∣m(−1)mdϕ(d)+∑d∣tϕ(2d)+…+∑d∣tpfi(d)=−m. If m is even, let m=2kt for odd t and k≥1. Then  ∑d∣m(−1)mdϕ(d)=∑d∣tϕ(d)+∑d∣tϕ(2d)+…+∑d∣tϕ(2k−1d)−∑d∣tϕ(2kd)  =(1+ϕ(2)+…+ϕ(2k−1)−ϕ(2k))∑d∣tϕ(d)=0  Hence the sum equals to

Tail3vn · Accepted Answer

Another approach is to use Lambert Series generating function:  q(1−q)2=∑n=1∞ϕ(n)qn1−qn=∑n=1∞ϕ(n)q−n−1|q∣&amp;lt;1⇒  q2(1−q2)2=∑n=1∞ϕ(n)q−2n−1=12∑n=1∞[ϕ(n)q−n−1−ϕ(n)q−n+1]  Hence:  ∑n=1∞ϕ(n)q−n+1=∑n=1∞ϕ(n)q−n−1−2∑n=1∞ϕ(n)q−2n−1  =q(1−q)2−2q2(1−q2)2=q1+q2(1−q2)2

Evaluate \sum_{n=1}^\infty\frac{\phi(n)}{7^n+1}

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