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

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

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!

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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).

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

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

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

therefore,

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

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

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