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Let $$S_n=\sum_{i=1}^{n}\sin k,\quad S_0=0.$$

Then

$$\sum_{k=1}^{n}\frac{\sin k}{k}=\sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k}=\frac{S_n}{n}+\sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}.$$

Could someone please explain the steps in getting this last equality. I am told it is using Abel's summation, but I have been reading up on that and I still can't see it.

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Let $a(n)$ and $f(n)$ be sequences and $A(n) = \displaystyle \sum_{k=1}^{n} a(k)$.

Abel's partial summation technique: \begin{align*} \sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\ & = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\ & = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n)) \end{align*} (The above is nothing but the discrete version of integration by parts).

$$\sum_{n=1}^{N} a(n) f(n) = \int_{1^-}^{N^+} f(t) d(A(t)) = f(t) A(t) \rvert_{1^-}^{N^+} - \int_{1^-}^{N^+} A(t) f'(t) dt$$ (The second integral can be interpreted as a Riemann-Stieltjes integral.)

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Set

$$S_n=\sum_{k=1}^n \sin k \;\; S_0=0$$

then

$$\sin k=\sum_{i=1}^{k}\sin i -\sum_{i=1}^{k-1}\sin i=S_k-S_{k-1}$$

Thus

$$\sum_{k=1}^{n}\frac{\sin k}{k}=\sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k}$$

We aim to show this is $$=\frac{S_n}{n}+\sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}$$

Set $a_k=\frac{1}{k}$ Then

$$\frac 1 {k(k+1)}=-\left(\frac{1}{k+1}-\frac 1 {k}\right)=a_{k+1}-a_k$$

So the formula in question is

$$={S_n}a_n-\sum_{k=1}^{n-1}{S_k}(a_{k+1}-a_k)$$

which is indeed Abel's summation by parts

$$\sum_{k=0}^na_kb_k=b_nA_n-\sum_{k=0}^{n-1} A_k(b_{k+1}-b_k)$$

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