Evaluate ∞∑n=1ϕ(n)7n+1
where ϕ(n) is Euler's totient function.
I found this problem on a Discord server I just joined. The first time I saw this sum, I was daunted. After gathering courage, I started my work on this problem.
I tried various things, like using the definition
ϕ(n)=n(1−1p1)⋯(1−1pm) (where the pi are the prime factors of n) and what not, but it lead to nothing.
Then I started thinking about the 7 in the denominator; why only 7? What is so special about 7? What is the difference between
∞∑n=1ϕ(n)6n+1 and
∞∑n=1ϕ(n)7n+1.
To answer that question, I took the literal difference of the two sums, but the denominator blew up so I started to think again. Because of that, I defined the two sequences and their sum:
an=ϕ(n)7n+1 and bn=ϕ(n)7n−1
S1=∞∑n=1an and S2=∞∑n=1bn
and took THEIR difference, which lead to
∞∑n=1(bn−an)=2∞∑n=1anbnϕ(n)
Now I can say that I'm genuinely stuck. Can someone provide a hint as to what to do next?
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