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Compute the sum \(\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots\)

michaelcai  Oct 5, 2017
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 #1
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      6  

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6.9.12

Guest Oct 10, 2017
 #2
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Let's define:

\(A(n)=\frac{2}{n*(n+1)*(n+2)}\\ S(n)=A(1)+A(2)+A(3)+.....+A(n)=\sum_{i=1}^{n}A(i)=\sum_{i=1}^{n}\frac{2}{i*(i+1)*(i+2)}\\\)

Now i am going to try and find a pattern for S(n) (a pattern that describes S(n) for every n):

 

 

\(S(1)=\frac{2}{1*2*3}=\frac{2}{6}\\ S(2)=\frac{2}{1*2*3}+\frac{2}{2*3*4}=\frac{5}{12}\\ S(3)=\frac{2}{1*2*3}+\frac{2}{2*3*4}+\frac{2}{3*4*5}=\frac{9}{20}\\ S(4)=\frac{2}{1*2*3}+\frac{2}{2*3*4}+\frac{2}{3*4*5}+\frac{2}{4*5*6}=\frac{14}{30}\)

 

The results seem to get closer and closer to one half...Let's investigate!

 

Let's look at the denominators of the fractions: 6, 12, 20, 30. The bolded numbers are the differences of the denominators-12=6+6, 20=12+8, 30=20+10.

It seems that the differences between the denominators keep growing! 6, then 8, then 10.... The differneces of the differences of the denominators seem to grow by 2! Maybe it's an Arithmetic progression???

 

Now, before anyone posts a long, bitter response to this answer, I am NOT saying that the pattern i found is a pattern that TRULY describes S(n), and there are other ways to solve this question without having to guess what Some of the formulas are. However, sometimes trying to find patterns can be very usefull- when you find one you can test it's validity using induction.

 

ANOTHER thing i want to add, is that i chose the denominator of the fraction for a reason. I could have written 7/15 instead of 14/30. why did i do that? what's so special about those denominators? I chose the denominators so that every fraction will look like this: \(\frac{x-1}{2x}\)

(When x is some natural number)

 

I know that it helps me understand that the series converges to one half, and helps me find a pattern to the denominators, meaning that i also have a pattern for the numerators.

 

Now I am going to guess what S(n) is using a fraction (like the previous sums i showed). Now i will try and find the denominator- The first denominators were 6, 12, 20 and 30- and i suggested that the differences between the demoninators grow by 2 with every new denominators (suppose the nth denominator is Dn, then Dn-Dn-1=2+2n). We know that D1=6, now let's find the nth denominator-

 

nth denominator=Dn=(Dn-Dn-1)+(Dn-1-Dn-2)+.......+(D2-D1)+D1=(2+2n)+(2+2(n-1))+......+(2+2*2)+6=6+(n-1)*(n+4).

 

 

That means the nth sum is going to have 6+(n-1)*(n+4) as the denominator. I also mentioned that i wanted the fraction to look like this: \(\frac{x-1}{2x}\)

 

 

This means that i am going to ASSUME that the nth numerator is going to be:
\(\frac{6+(n-1)*(n+4)}{2}-1=\frac{4+(n-1)*(n+4)}{2}\)

 

 

SO, the fraction that represents S(n) is going to be:

\(\frac{\frac{4+(n-1)*(n+4)}{2}}{6+(n-1)*(n+4)}=\frac{4+(n-1)*(n+4)}{2*(6+(n-1)*(n+4))}\)

 

 

Will continue soon- have to go

 

~blarney master~

Guest Oct 11, 2017
 #3
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continuing my answer:

 

Now, after i guessed that

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

 

I will prove it using induction- Suppose \(S(n)=\frac{n*(n+3)}{2*(n+1)*(n+2)}\)

I will prove that \(S(n+1)=\frac{(n+1)*(n+4)}{2*(n+2)*(n+3)}\):

 

\(S(n+1)=\sum_{i=1}^{n+1}A(i)=(\sum_{i=1}^{n}A(i))+A(n+1)=\\ S(n)+A(n+1)=\frac{n*(n+3)}{2*(n+1)*(n+2)}+\frac{2}{(n+1)*(n+2)(n+3)}=\\ \frac{4+n*(n+3)^2}{2*(n+1)*(n+2)(n+3)}=\frac{(n+1)^2*(n+4)}{2*(n+1)*(n+2)(n+3)}=\\ \frac{(n+1)*(n+4)}{2*(n+2)(n+3)}\)

Now we just need to prove that it works for n=1:

 

\(S(1)=\sum_{i=1}^{1}A(i)=\sum_{i=1}^{i}\frac{2}{i*(i+1)(i+2)}=A(i)=\frac{2}{1*2*3}=\frac{1}{3}=\\ \frac{1*(1+3)}{2*(1+1)*(1+2)}\)

 

Now we finally know that \(S(n)=\frac{n*(n+3)}{2*(n+1)*(n+2)}\).

 

Now all you need to do is prove that when n goes to infinity S(n) goes to 1/2.

 

Good luck!

 

~blarney master~

Guest Oct 11, 2017

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