What is the sum of the infinite series 1/[n(n+1)(n+2)]?

[adamg]

hey, i was just wondering if anyone could help find the sum of the infinite series defined by 1/[n(n+1)(n+2)]. I can split it into partial fractions but not sure from there. Thanks

 

[strid]

its a/(1-r)...
where |r| < 1

where a is the initial values (1/6), and r is the common ratio...
it to late for my brain to think right now so I can tthink of what the ratio is going to be.. wil try tomorrow if nobody else has helped you.. going tp sleep now..gonight!

 

[dextercioby]

I'm not sure the splitting can do you any good,because each of the series you'd be geting would be infinite.If you don't know the answer,well,my Maple said it's 1/4...i don't know how it did it...

I'm going to check it in G & R,too...

Daniel.

 

[adamg]

i thought you could do it by splitting it into partial fractions, then finding an expression for the nth partial sum and let n tend to infinity

 

[dextercioby]

According to Maple,the partial sum is:
$$\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)} =-\frac{1}{2(n+2)(n+1)}+\frac{1}{4}$$

If you can find a way to prove what I've just written,then that's it...

Daniel.

 

[Hurkyl]

I think partial fractions is promising too -- with care, you can probably arrange the terms into one or more telescoping series. What did you get when you applied partial fractions?

 

[dextercioby]

Here's the proof for the partial sum:

$$S=:\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}$$(1)
-----------------------------||------------------------

Use partial fractions to rewrite the initial partial sum as:

$$S=\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{k+1}+\sum_{k=1}^{n}\frac{1}{2(k+2)}$$ (2)

According to Maple each of the three sums is:

$$\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma$$ (3)

$$\sum_{k=1}^{n} \frac{-1}{k+1}=-\psi(n+2)+1-\gamma$$ (4)

$$\sum_{k=2}^{n} \frac{1}{2(k+2)}=\frac{1}{2}\psi(n+3)-\frac{3}{4}+\frac{1}{2}\gamma$$(5)

Add (3)->(5) and equate with (2):

$$S=\frac {1}{4}+\frac{1}{2}\psi(n+1)-\psi(n+2)+\frac{1}{2}\psi(n+3)$$(6)

Now,use the property of "psi":

$$\psi(z+1)=\psi(z)+\frac{1}{z}$$ (7) for $z\neq 0$

to write:

$$\psi(n+2)=\psi(n+1)+\frac{1}{n+1}$$ (8)

$$\psi(n+3)=\psi(n+2)+\frac{1}{n+2}=\psi(n+1)+\frac{1}{n+1}+\frac{1}{n+2}$$ (9)

to rewrite the "3-psi" term from (6) simply as:

$$-\frac{1}{2(n+1)(n+2)}$$ (10)

Then add (10) to 1/4 from (6) to get the equality you were supposed to prove...

In the initial equality set $n\rightarrow +\infty$ to get the answer $\frac{1}{4}$

Daniel.

 

[Hurkyl]

What do you think of this decomposition?

$$\frac{1}{k (k+1) (k+2)} = \frac{1}{2}<br /> \left(<br /> \frac{1}{k} - \frac{1}{k+1} - \frac{1}{k+1} + \frac{1}{k+2}<br /> \right)$$

 

[dextercioby]

That one gives the answer,too...1/4...

Daniel.

 

[adamg]

What do you think of this decomposition?

$$\frac{1}{k (k+1) (k+2)} = \frac{1}{2}<br /> \left(<br /> \frac{1}{k} - \frac{1}{k+1} - \frac{1}{k+1} + \frac{1}{k+2}<br /> \right)$$

i got this far, then not sure how to cancel terms to find the sum?

 

[dextercioby]

Well,it's not difficult.You may write it as
$$\frac{1}{2}[(\frac{1}{k}-\frac{1}{k+1})-(\frac{1}{k+1}-\frac{1}{k+2})]$$

and then give values to "k" to see the pattern the terms take...

Daniel.

 

[adamg]

yeah got it now, thanks for all the help!

 

[saltydog]

According to Maple each of the three sums is:

$$\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma$$ (3)

$$\sum_{k=1}^{n} \frac{-1}{k+1}=-\psi(n+2)+1-\gamma$$ (4)

$$\sum_{k=2}^{n} \frac{1}{2(k+2)}=\frac{1}{2}\psi(n+3)-\frac{3}{4}+\frac{1}{2}\gamma$$(5)

Daniel.

Thanks, didn't have a clue. Turns out the first one is the following:

$$\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\{\frac{\int_0^\infty e^{-t}t^{n}\ln(t)dt}{\int_0^\infty e^{-t}t^{n}dt}+\int_0^\infty \frac{1}{t}(\frac{1}{1+t}-e^{-t})dt\}$$

I had no idea! Does Adamg know this? Don't mind me, I mean I had problems with Leibnitz's rule the other day. I tell you what though, if I was the teacher assigning this problem, we'd be looking at these integrals. Now how in the world does the sum turn out to be this messy expression?

 

[dextercioby]

It comes from the logarithmic derivative of the Gamma Euler function,which is "psi"...

Daniel.

 

[TRUSKY1965]

Is an easy sum:

$$\sum\frac{1}{n(n+1)(n+2)}$$=$$\sum\frac{1}{2n}$$+$$\frac{1}{2(n+2)}$$-$$\frac{1}{n+1}$$= ($$\frac{1}{2}$$+$$\frac{1}{4}$$+$$\frac{1}{6}$$+$$\frac{1}{8}$$+$$\frac{1}{10}$$...+$$\frac{1}{6}$$+$$\frac{1}{8}$$+$$\frac{1}{10}$$+...+(-$$\frac{1}{2}$$-$$\frac{1}{3}$$-$$\frac{1}{4}$$-$$\frac{1}{5}$$-$$\frac{1}{6}$$-...) = $$\frac{1}{2}$$+$$\frac{1}{4}$$+$$\frac{1}{3}$$+$$\frac{1}{4}$$+$$\frac{1}{5}$$+... +(-$$\frac{1}{2}$$-$$\frac{1}{3}$$-$$\frac{1}{4}$$-$$\frac{1}{5}$$-$$\frac{1}{6}$$-...) = $$\frac{1}{4}$$.