6
$\begingroup$

$$ \frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots $$ 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.

$\endgroup$
  • $\begingroup$ 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 :) $\endgroup$ – user230452 Feb 19 '16 at 11:42
  • $\begingroup$ 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. $\endgroup$ – user230452 Feb 19 '16 at 11:43
  • $\begingroup$ Related : math.stackexchange.com/questions/560816/… $\endgroup$ – lab bhattacharjee Feb 19 '16 at 15:41
5
$\begingroup$

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: $$\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}.$$

$\endgroup$
  • 1
    $\begingroup$ Ah, much easier, and probably generalizes to handle $\sum_{n=k}^{\infty} \frac{1}{\binom{n}{k}}$. $\endgroup$ – lhf Feb 19 '16 at 11:57
4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.