$$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$$=$$\frac{1}{2} \left(-\frac{2}{n+1}+\frac{1}{n+2}+\frac{1}{n}\right)$$

$$s_n=\frac{1}{2} \left(\left(\frac{1}{n+2}+\frac{1}{n}-\frac{2}{n+1}\right)+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)+\ldots +\left(1 \frac{1}{1}-\frac{2}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{3}+\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{2}{7}+\frac{1}{8}\right)\right)$$

Im sorry for not being able to get the term in right order, but as you can see the first two terms in the parenthesis should be the last for a accurate representation.

$$\left( \begin{array}{cc} 1 & 1-1+\frac{1}{3} \\ 2 & \frac{1}{2}-\frac{2}{3}+\frac{1}{4} \\ 3 & \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \\ 4 & \frac{1}{4}-\frac{2}{5}+\frac{1}{6} \\ 5 & \frac{1}{5}-\frac{1}{3}+\frac{1}{7} \\ 6 & \frac{1}{6}-\frac{2}{7}+\frac{1}{8} \\ \end{array} \right)$$

I think that my pure understanding of the representation is what makes me confused. From the table I see that a pattern in the denominators emerge, when n=1, then (1-2+3), when n=2 then, 2-3+4. However given the part of sn where $$\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}$$ I fail to see how the first part of expression right above is not undefined when n=1.

I am convinced that it is my lack of knowledge of what the representation actually means. And I am also struggling to see what cancels to give$$\frac{1}{2} \left(\frac{1}{n+2}-\frac{1}{n+1}+\frac{1}{2}\right)$$

However given that this is true I am able to understand that the answer is (1/4). But as I said it is the representation I do not get. Could someone help me as this would be very useful for my next chapter in my textbook!


You should write: (define $s_n$ as the sequence of partial sums of the series)

$s_n=\sum _{k=1}^{n} \frac{1}{k (k+1) (k+2)}$ $=\sum _{k=1}^{n} {\frac{1}{2} \left(-\frac{2}{k+1}+\frac{1}{k+2}+\frac{1}{k}\right)}$

$=\frac{1}{2}\left(-2\sum_{k=1}^{n}\frac{1}{k+1}+\sum_{k=1}^{n}\frac{1}{k+2}+\sum_{k=1}^{n}\frac{1}{k}\right)$ because the sums are finite. Observe then that the three sums overlap on most terms.

$=\frac{1}{2}\left(-2\sum_{k=2}^{n+1}\frac{1}{k}+\sum_{k=3}^{n+2}\frac{1}{k}+\sum_{k=1}^{n}\frac{1}{k}\right)$ by just rewriting the sum.

$=\frac{1}{2}\left(\left(-2\sum_{k=3}^{n}\frac{1}{k}-2\cdot\frac{1}{2}-2\frac{1}{n+1}\right)+\left(\sum_{k=3}^{n}\frac{1}{k}+\frac{1}{n+1}+\frac{1}{n+2}\right)+\left(\sum_{k=3}^{n}\frac{1}{k}+\frac{1}{1}+\frac{1}{2}\right)\right)$ by just rewriting the sum again.

$=\frac{1}{2}\left(-2\cdot\frac{1}{2}-2\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{1}+\frac{1}{2}\right)$ by cancelling terms.

$=\frac{1}{2}\left(-\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{2}\right)$ by cancelling terms.

Then only now, can you consider taking the limit $n\to\infty$.

Maybe you are not understanding the $\sum$ operator.

$\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$ is exactly $\frac{1}{\underline{1}(\underline{1}+1)(\underline{1}+2)}+\frac{1}{\underline{2}(\underline{2}+1)(\underline{2}+2)}+\dots+\frac{1}{\underline{n}(\underline{n}+1)(\underline{n}+2)}$ by definition.

And $\sum_{k=1}^{\infty} \frac{1}{k(k+1)(k+2)}$ is exactly $\lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$ by definition.

More Examples

$$\sum_{k=1}^n a_{k+1}=a_2+a_3+\dots+a_{n+1}=\sum_{k=2}^{n+1} a_k \quad (Eq1)$$

In general, we have $\sum_{x\in S} f(x)$, which means "take all the distinct values in a finite set $S$, apply $f$ to each of them, and sum the results up". Note the results may not be distinct.

Now we can write $\sum_{k=1}^n f(k)=\sum_{k\in\mathbb{Z}\cap[1,n]} f(k)$.

If $g$ is an injective function on $S$ (distinct inputs give distinct outputs), then $$\sum_{x\in S} f(x)=\sum_{y\in g(S)} f(g^{-1}(y))\quad\text{, also}\quad\sum_{x\in S} f(g(x))=\sum_{y\in g(S)} f(y)$$ where $g(S)$ are all the outputs of $g$ given inputs in $S$, and $g^{-1}$ is the function which takes as input an output of $g$ and returns the input which gave that output (this is the inverse function).

So in (Eq1), we are taking $g$ to be $k\mapsto k+1$, and $S=\mathbb{Z}\cap[1,n]$. Hence $g(S)=\mathbb{Z}\cap[2,n+1]$.

Of course, you have the distributive rule: $\sum_{x\in S}af(x)=a\sum_{x\in S}f(x)$ (use induction).

  • $\begingroup$ I see that I am missing some knowledge of how to manipulate sum. Would you be able to give me some guidance in the bottom of the post? $\endgroup$ – ALEXANDER Jan 18 '15 at 18:01
  • $\begingroup$ @ALEXANDER OK, I've added a section; see if you can understand the general concept. $\endgroup$ – user1537366 Jan 19 '15 at 5:04
  • $\begingroup$ Perfect I got it! $\endgroup$ – ALEXANDER Jan 19 '15 at 14:55

Hint: We are finding $$\sum_1^\infty \frac{1}{2}\left(\left(\frac{1}{n}-\frac{1}{n+1}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right) \right).$$

  • $\begingroup$ Could you explain a bit further? $\endgroup$ – ALEXANDER Jan 18 '15 at 18:57
  • $\begingroup$ Forget about the $\frac{1}{2}$ in front temporarily. The sum $\sum_1^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)$ is telescoping, sum $1$. The second sum is also telescoping, only the first term $\frac{1}{2}$ survives. So (apart from the $\frac{1}{2}$ in front, the sum is $1-\frac{1}{2}=\frac{1}{2}$. Finally, multiply by the $\frac{1}{2}$ in front. We get $\frac{1}{4}$. $\endgroup$ – André Nicolas Jan 18 '15 at 19:05

Your diagram is perfect. All these cancel.. It's a telescoping sum enter image description here


$\dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2n(n+1)} - \dfrac{1}{2(n+1)(n+2)}$


$$\sum_{n = 1}^{\infty} \dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2*1*2} = \dfrac{1}{4}$$

we used the telescoping series formula $\sum_{n=1}^\infty (f(n) - f(n+1)) = f(1) - \lim_{n \to \infty} f(n)$


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.