# Finding the sum to n terms of series :$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$

$$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$$ up to $n$ terms. I need help in solving this sum. I tried finding the coefficients of terms after splitting the terms..: it becomes $$(\frac{1}{1\cdot 6}-\frac{1}{2\cdot 2}+\frac{1}{2\cdot 3}-\frac{1}{6\cdot4}) + (\frac{1}{6\cdot 2} - \frac{1}{3\cdot2} +\frac{1}{4\cdot 2} -\frac{1}{6\cdot5})+\cdots.$$ I tried solving it but am getting nowhere .Someone please help me with this sum.

• The general tactic in these kind of sums where the denominator is the product of terms suspcisouly doffering by one is to split as partial fractions and then telescope :) – Saikat Feb 19 '16 at 11:42
• You can telescope when they differ by something other than 1 too. It is just that when they differ by 1, it is easy to split them into partial fractions and intuitively guess their coefficients. – Saikat Feb 19 '16 at 11:43
• – lab bhattacharjee Feb 19 '16 at 15:41

## 3 Answers

We can use the following identity: $$\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left(\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right).$$ Thanks to this identity, if we want to compute $$\sum_{k=1}^n\frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3}\sum_{k=1}^n\left(\frac{1}{k(k+1)(k+2)}-\frac{1}{(k+1)(k+2)(k+3)}\right),$$ we only have to subtract $1/(n+1)(n+2)(n+3)$ from $1/(1\cdot2\cdot3)$ and then devide it by 3, because all other terms cancel out. This gives us the result: $$\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}.$$

• Ah, much easier, and probably generalizes to handle $\sum_{n=k}^{\infty} \frac{1}{\binom{n}{k}}$. – lhf Feb 19 '16 at 11:57

Hint: Use partial fractions: $$\frac1{(n-3)(n-2)(n-1)n}=-\frac1{2(n-2)}+\frac1{2(n-1)}-\frac1{6n}+\frac1{6(n-3)}$$ and note that it telescopes, so that you can find the partial sums.

HINT:

The $r(\ge1)$th term

$$=\dfrac1{r(r+1)(r+2)(r+3)} =\dfrac{r+3-r}{3r(r+1)(r+2)(r+3)}=v_r-v_{r+1}$$

where $v_m=\dfrac1{3m(m+1)(m+2)}$

$$\sum_{r=1}^n u_r=\sum_{r=1}^n(v_r-v_{r+1})=v_1-v_{n+1}$$

Can you take it from here?