Prove that $\lim\limits_{n\to+\infty}\sum\limits_{i=1}^n\prod\limits_{j=1}^i\frac{3j}{3j+4}=3$

Question

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.

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Input to Mathematica:

Limit[Sum[Product[(3 j)/(3 j + 4), {j, 1, i}], {i, 1, n}], 
 n -> +\[Infinity]]

and the output was $\frac{7~ \Gamma \left(7/3\right)}{\Gamma \left(10/3\right)}=3$.

Answer 1

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!

Answer 2

Another approach using a generating function and using only elementary functions.

Giving a name to the products involved,

$$ P_k = \prod_{j=1}^k \frac{3j}{3j+4} $$

These have the recurrence relation

$$ P_k = \frac{3k}{3k+4} P_{k-1} $$

when $k \geq 1$, and $P_0=1$.

Define the generating function

$$ f(x) = \sum_{k=0}^\infty P_k x^k $$

Note that if $x>0$ and $f(x)$ converges, $f(x)>0$; and if $x>0$ and $f'(x)$ exists, $f'(x)>0$.

The usual generating function manipulations give

$$ \begin{align*} x f'(x) &= \sum_{k=0}^\infty k P_k x^k \\ x^2 f'(x) + x f(x) &= \sum_{k=1}^\infty k P_{k-1} x^k \end{align*} $$

Therefore

$$ \begin{align*} 3x f'(x) + 4 f(x) - 3(x^2 f'(x) + x f(x)) &= \sum_{k=0}^\infty (3k+4) P_k x^k - \sum_{k=1}^\infty 3k P_{k-1} x^k \\ (3x-3x^2) f'(x) + (4-3x)f(x) &= 4 P_0 + \sum_{k=1}^\infty ((3k+4) P_k - 3k P_{k-1}) x^k \\ 3x(1-x) f'(x) + (4-3x) f(x) &= 4 \end{align*} $$

The general solution to

$$3x(1-x) g'(x) + (4-3x) g(x) = 0$$

is found by

$$ \frac{g'(x)}{g(x)} = \frac{3x-4}{3x(1-x)} = -\frac 13 \left(\frac{1}{1-x} + \frac{4}{x}\right) $$ $$ \ln |g(x)| = \frac 13 \ln|1-x| - \frac 43 \ln|x| + c_1 $$ $$ g(x) = c_2 x^{-4/3} (1-x)^{1/3} $$

For the particular solution, suppose $f(x) = g(x) h(x)$ where $g(x) = x^{-4/3} (1-x)^{1/3} $ satisfies $3x(1-x) g'(x) + (4-x) g(x) = 0$. The differential equation for $f$ is then

$$ 3x(1-x) f'(x) + (4-x) f(x) = 4 $$ $$ [3x(1-x) g'(x) + (4-x) g(x)] h(x) + 3x(1-x) g(x) h'(x) = 4 $$ $$ 3 x^{-1/3} (1-x)^{4/3} h'(x) = 4 $$ $$ h'(x) = \frac 43 x^{1/3} (1-x)^{-4/3} $$ $$ f(x) = x^{-4/3} (1-x)^{1/3} \left[C + \frac 43 \int_{\frac 12}^x t^{1/3}(1-t)^{-4/3}\, dt\right] $$ for some constant $C$.

If $0<t<x<1$, then

$$ \begin{align*} t^{1/3} (1-t)^{-4/3} &< x^{1/3} (1-t)^{-4/3} \\ \int_{\frac 12}^x t^{1/3} (1-t)^{-4/3}\, dt &\leq x^{1/3} \int_{\frac 12}^x (1-t)^{-4/3}\, dt \\ \int_{\frac 12}^x t^{1/3} (1-t)^{-4/3}\, dt &\leq 3 x^{1/3} \left((1-x)^{-1/3} - 2^{1/3} \right) \\ f(x) &\leq C x^{-4/3} (1-x)^{1/3} + \frac{4}{x} \left(1 - (2-2x)^{1/3}\right) \end{align*} $$

Since on $(0,1)$, $f$ is increasing and continuous and bounded above by a continuous function which has a limit as $x \to 1^-$, $\lim_{x \to 1^-} f(x)$ exists. Evaluating it,

$$ \begin{align*} \lim_{x \to 1^-} f(x) &= \lim_{x \to 1^-} x^{-4/3} (1-x)^{1/3} \left[C + \frac 43 \int_{\frac 12}^x t^{1/3} (1-t)^{-4/3} dt\right] \\ \lim_{x \to 1^-} f(x) &= 1^{-4/3} \lim_{x \to 1^-} \frac{1}{(1-x)^{-1/3}} \left[C + \frac 43 \int_{\frac 12}^x \frac{t^{1/3}} {(1-t)^{4/3}} dt\right] \\ \lim_{x \to 1^-} f(x) &= \lim_{x \to 1^-} \frac{1}{\frac 13 (1-x)^{-4/3}} \cdot \frac{4x^{1/3}}{3(1-x)^{4/3}} \\ \lim_{x \to 1^-} f(x) &= 4 \\ \sum_{k=0}^\infty P_k 1^k &= 4 \\ 1 + \sum_{k=1}^\infty P_k &= 4 \end{align*} $$

So as claimed,

$$ \sum_{i=1}^\infty P_i = \lim_{n \to \infty} \sum_{i=1}^n \prod_{j=1}^i \frac{3j}{3j+4} = 3 $$