Prove that $\lim\limits_{n\to+\infty}\sum\limits_{i=1}^n\prod\limits_{j=1}^i\frac{3j}{3j+4}=3$.
I can't think of any method to prove it. But here's how I made this problem.
Firstly we know $\lim\limits_{n\to+\infty}\sum\limits_{i=1}^n\frac1n$ does not exist. But what if we consider something smaller? Since $\frac1n=\prod\limits_{i=1}^{n-1}\frac i{i+1}$, and $\frac{3i}{3i+4}<\frac{3i}{3i+3}=\frac i{i+1}$, I came up with the problem at the beginning.
Input to Mathematica:
Limit[Sum[Product[(3 j)/(3 j + 4), {j, 1, i}], {i, 1, n}],
n -> +\[Infinity]]
and the output was $\frac{7~ \Gamma \left(7/3\right)}{\Gamma \left(10/3\right)}=3$.
Best Answer
First, recall the definition of the hypergeometric function: $$ \begin{aligned} {}_2F_1(a,b;c;z) &= \sum_{n\ge 0} \frac{(a)_n(b)_n}{n!(c)_n} \\ &= 1 +\frac{a\cdot b}{1\cdot c}z +\frac{a(a+1)\cdot b(b+1)}{1.2\cdot c(c+1)}z^2 \\ &\qquad+\frac{a(a+1)(a+2)\cdot b(b+1)(b+2)}{1.2.3\cdot c(c+1)(c+2)}z^3 \\ &\qquad\qquad+\frac{a(a+1)(a+2)(a+3)\cdot b(b+1)(b+2)(b+3)}{1.2.3.4\cdot c(c+1)(c+2)(c+3)}z^4 \\ &\qquad\qquad\qquad+\dots\ . \\[3mm] &\qquad\text{ And we want to compute the following sum $S$:} \\[3mm] S &=\sum_{n\ge 1} \frac{1.2.\dots.n}{\left(\frac 73\right)\left(\frac 73+1\right)\dots\left(\frac 73+(n-1)\right)}\\ &={}_2F_1\left(1,1;\frac 73;1\right)-1=4-1=3\ . \\[3mm] &\qquad\text{ because of} \\[3mm] F_1\left(1,1;\frac 73;1\right) &= \frac {\Gamma\left(\frac73\right)\Gamma\left(\frac73-1-1\right)} {\Gamma\left(\frac73-1\right)\Gamma\left(\frac73-1\right)} =\frac{4/3}{1/3} \\ &=\bbox[yellow]{\qquad 4\qquad}\ . \end{aligned} $$ Here we have used the special values formula at $z=1$.
And $\Gamma(z+1)=z\Gamma(z)$ for $z$ equal to $4/3$ and $4/3-1$.
Feel free to generalize!