# Is there a way to simplify this summation further?

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$$\varphi(5,n)=1+2\sum\limits_{k=1}^n\frac 1{(5k)^3-5k}$$

## 1 Answer

I do not know how much this could help you.

If you want to use harmonic numbers $$\sum\limits_{k=1}^{4n+1}\frac 1{n+k}=H_{5 n+1}-H_n$$ $$\sum\limits_{k=1}^{n}\frac 1{k}=H_n$$ Now, for the other terms, you can write them using $$\sum\limits_{k=1}^n\frac 1{ak+b}=\frac 1a\sum\limits_{k=1}^n\frac 1{k+\frac ba}=\frac 1a\left( H_{n+\frac ba}-H_{\frac ba}\right)$$

• So there isn't really a way to simplify the summation... – Crescendo May 31 '17 at 15:07