Calculate $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$

Hint: $$\frac{1}{n(n+1)(n+2)}=\frac{1/2}{n(n+1)}-\frac{1/2}{(n+1)(n+2)}$$

Consider

$$f(x) = \sum_{n=1}^{\infty} \frac{x^{n+2}}{n (n+1)(n+2)} $$

Then

$$f''(x) = -\log{(1-x)}$$

$$f'(x) = (1-x) \log{(1-x)} +x $$

$$f(x) = -\frac14 [x (2-x) - 2 (1-x)^2 \log{(1-x)}] + \frac12 x^2$$

The sum is then $f(1) = 1/4$.

Your partial sums are the same as those of $\frac 1 2 H_n-H_{n+1}+\frac 1 2 H_{n+2}$ minus some terms. Find those terms and use those harmonic sums tend to 0 in the limit. The answer is $1/4$. More generally, it happens that $$\binom{n+k-1}{k}^{-1}=\frac{k!}{n(n+1)\cdots (n+k)}=\sum_{j=0}^k \binom kj \frac{(-1)^j}{n+j}$$

This means that $$\sum_{n=1}^N\binom{n+k-1}k^{-1}=\sum_{j=0}^k\binom kj(-1)^j (H_{N+j}-H_j)$$

Using this and the fact that $\sum_{j=0}^k\binom kj(-1)^{j+1} H_j=\frac 1 k$ one gets $$\sum_{n\geqslant 1}\binom {n+k-1}k^{-1}=\frac 1 k$$

or $$\sum_{n\geqslant 1}\frac {1}{n(n+1)\cdots (n+k)}=\frac 1k\frac 1 {k!}$$