I've been doing a bit of summation work, and I got to$$\sum\limits_{k=1}^{4n+1}\frac 1{n+k}-\sum\limits_{k=1}^n\frac 1{5k-3}-\sum\limits_{k=1}^{n}\frac 1{5k-2}+\frac 35\sum\limits_{k=1}^n\frac 1k\tag1$$And I'm wondering if there is a way to further simplify $(1)$. Specifically, the last three terms, because I feel that you can somehow simplify them even more.
I tried expanding the terms out and got$$\begin{align*} & \sum\limits_{k=1}^n\frac 1{5k-3}+\sum\limits_{k=1}^n\frac 1{5k-2}-\frac 35\sum\limits_{k=1}^n\frac 1k\\ & =\left(\frac 12+\frac 17+\cdots+\frac 1{5n-3}\right)+\left(\frac 13+\frac 18+\cdots+\frac 1{5n-2}\right)-\left(\frac 35+\frac 3{10}+\frac 3{15}+\cdots+\frac 3{5n}\right)\\ & =\left(\frac 12+\frac 13+\frac 17+\frac 18+\cdots+\frac 1{5n-3}+\frac 1{5n-2}\right)-\frac 35\sum\limits_{k=1}^n\frac 1k\end{align*}$$And that's how far I got. Any ideas?
Just as clarification, I started with$$\varphi(5,n)=1+2\sum\limits_{k=1}^n\frac 1{(5k)^3-5k}$$