Sum of the following series

Question

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.

Answer 1

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 $$ \begin{align} \frac{\Gamma(\frac23)}{\Gamma(\frac13)}\sum_{k=1}^n\frac{\Gamma(k+\frac13)}{\Gamma(k+\frac23)} &=\frac{\Gamma(\frac23)}{\Gamma(\frac13)}\frac32\left[\frac{\Gamma(n+\frac43)}{\Gamma(n+\frac23)}-\frac{\Gamma(\frac43)}{\Gamma(\frac23)}\right]\\[6pt] &=-\frac12+\frac12\frac{\Gamma(\frac23)}{\Gamma(\frac43)}\frac{\Gamma(n+\frac43)}{\Gamma(n+\frac23)}\\[6pt] &=-\frac12+\frac12\frac{\frac43\cdot\frac73\cdots(n+\frac13)}{\frac23\cdot\frac53\cdots(n-\frac13)}\\[6pt] &=-\frac12+\frac12\frac{4\cdot7\cdots(3n+1)}{2\cdot5\cdots(3n-1)}\tag{3} \end{align} $$

Answer 2

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: $$\sum_{k=1}^{n}A_k = \frac{B_n-2}{2}.$$

Answer 3

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....