# Sum of the following series

I'm stuck with this series:$$\frac12 + \frac{1\cdot 4}{2 \cdot 5} + \frac{1\cdot 4 \cdot 7}{2 \cdot 5 \cdot8} + \frac{1 \cdot 4 \cdot 7 \cdot 10}{2 \cdot 5 \cdot 8 \cdot 11}+\ldots$$ I cant even find the $n$-th th term here...
I have to prove that

$$\sum_{k=1}^n A_k = \frac12 \left[ \frac{4\cdot 7 \cdot 10 \cdot \ldots \cdot(3n+1)}{2 \cdot 5 \cdot 8 \cdot\ldots\cdot(3n -1)}\ -1 \right]$$

Here are the instructions given in the question:

write the $A_n$ term of this series
and express $A_{n+1}$ using $A_n$

find $C$ and $D$ such that
$f(n) = (Cn + D)A_{n+1}$ and
$f(n) - f(n -1) = A_n$

and after that i have to come up with that above answer...

P.S :- Im stuck in the step of writing the $A_n$ term. I don't know how to write a $n^{th}$ term for a series like this. I think I could manage if someone could help show me how to write the $n^{th}$ term.

• Do those decimals represent multiplication? Sep 4, 2014 at 16:58
• Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Sep 4, 2014 at 16:58
• yes. the decimals represent multiplications Sep 4, 2014 at 16:59
• uh... i have no thoughts of my own to how to start this one because i cant write the $U_r$ term in this series ... if anyone can help me with it... even that would be a huge help Sep 4, 2014 at 17:01
• Is $U_r = (\prod_{k=0}^{r}(1+3k))/(\prod_{k=0}^{r}(2+3k))$ not adequate?
– user169852
Sep 4, 2014 at 17:13

We can use the identity $$\frac1{x-y+1}\left[\frac{\Gamma(n+1+x)}{\Gamma(n+y)}-\frac{\Gamma(n+x)}{\Gamma(n-1+y)}\right]=\frac{\Gamma(n+x)}{\Gamma(n+y)}\tag{1}$$ to prove, by induction, $$\sum_{k=m}^n\frac{\Gamma(k+x)}{\Gamma(k+y)}=\frac1{x-y+1}\left[\frac{\Gamma(n+1+x)}{\Gamma(n+y)}-\frac{\Gamma(m+x)}{\Gamma(m-1+y)}\right]\tag{2}$$ The series in the question can be rewritten as \begin{align} \frac{\Gamma(\frac23)}{\Gamma(\frac13)}\sum_{k=1}^n\frac{\Gamma(k+\frac13)}{\Gamma(k+\frac23)} &=\frac{\Gamma(\frac23)}{\Gamma(\frac13)}\frac32\left[\frac{\Gamma(n+\frac43)}{\Gamma(n+\frac23)}-\frac{\Gamma(\frac43)}{\Gamma(\frac23)}\right]\\[6pt] &=-\frac12+\frac12\frac{\Gamma(\frac23)}{\Gamma(\frac43)}\frac{\Gamma(n+\frac43)}{\Gamma(n+\frac23)}\\[6pt] &=-\frac12+\frac12\frac{\frac43\cdot\frac73\cdots(n+\frac13)}{\frac23\cdot\frac53\cdots(n-\frac13)}\\[6pt] &=-\frac12+\frac12\frac{4\cdot7\cdots(3n+1)}{2\cdot5\cdots(3n-1)}\tag{3} \end{align}

Notice that the $n$-th term of the sum is given by: $$A_n=\prod_{j=1}^{n}\frac{3j-2}{3j-1}.$$ Let now $B_n$ be: $$B_n = \prod_{j=1}^{n}\frac{3j+1}{3j-1}=(3n+1)\prod_{j=1}^{n}\frac{3j-2}{3j-1}=(3n+1)A_n=(3n-1)A_n+2A_n.$$ Since: $$(3n-1)A_n = (3n-2)A_{n-1} = B_{n-1}$$ it happens that: $$B_n = B_{n-1} + 2 A_n = B_{n-2} + 2 A_{n-1} + 2 A_n = \ldots =B_1 + 2\sum_{k=1}^{n}A_k.$$ Since $B_1 = 2$, the last identity gives: $$\sum_{k=1}^{n}A_k = \frac{B_n-2}{2}.$$

First of all I appreciate everyone's help with this question
But I dont understand the advanced answers given and this was just a question for .. i guess for beginner level.. which was in G.C.E A/L exam in our country - year 1989

The way that i think what the examiners may have expected to see should be something like this because our syllabus didn't include the above methods ( seems very advanced to me )

first i wrote

$A_{r+1}$ using $A_r$

$A_{r+1} = A_r\frac{3r +1}{3r +2}$

then i wrote f(n) as the questions asks

f(r)=$(Cr +D)A_{r+1}$

then i moved to the next step

$$f(r) - f(r-1) = A_r$$ $$(Cr +D)A_{r+1} - [C(r-1) +D]A_r= A_r$$ $$(Cr +D)A_r\frac{3r +1}{3r +2} - [C(r-1) +D]A_r= A_r$$

Now $A_r$ can be canceled out
then... I end up with this expression

$$(Cr +D)(3r +1) - (3r +2) [C(r-1) +D] = (3r +2)$$

Now considering the coefficient of r and the constant i got that

$c = \frac32$ and$D = 1$

$$f(r) - f(r-1) = A_n$$ $$r=1 ;f(1) - f(0) = A_1$$ $$r=2 ;f(2) - f(1) = A_2$$ $$r=3 ; f(3) - f(2) = A_3$$
If i add this whole thing until it goes to $$r=n ; f(n) - f(n-1) = A_n$$
I get

$$\sum_{r=1}^n A_r= f(n) - f(0)$$ $$\sum_{r=1}^n A_r= (3n +2)A_n\frac{(3n +1)}{2(3n +2)} - D(A_1)$$ $$\sum_{r=1}^n A_r=\frac12 [\frac{4.7.10....(3n +1)}{2.5.8....(3n -1)} - 1]$$

Thats the answer that i was able to come up with myself and i leave it up to u guys to check whether im wrong....