Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$ Consider the equation:
\begin{gather*}
\frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1
\end{gather*}
Prove that:

*

*For every $n \geq 2$, equation $(1)$ always has a unique positive solution $x_n$.


*The sequence $(x_n)$ has a finite limit with $n \to \infty+$. Find that limit.
Here's all I did:
$\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} = 1.$
$\Leftrightarrow 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) = n! (x+1)(x+2)...(x+n) $
Consider the polynomial $P(x) = 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) - n! (x+1)(x+2)...(x+n) $
Consider the highest coefficient of the polynomial:$ -n! < 0 $
Consider the lowest coefficient of the polynomial:$ (2.3...n)^2+(1.3...n)^2+...(1.2...(n-1))^2  - n! = (1.3...n)^2+...(1.2...(n-1))^2  > 0$
Let $x_1,x_2,x_3....$ be the solutions of$ P(x) $
Assume that equation (1) has no positive solution.
According to Viette's theorem: $x_1 x_2 x_3....x_n = (-1)^n \frac{a_0}{a_n} $
$\frac {a_0}{a_n} < 0 \Rightarrow $if $n$ is an odd number , then $x_1 x_2 x_3....x_n < 0$ , but
$(-1)^n \frac{a_0}{a_n} >0 \Rightarrow $ (nonsense) . Same with $n $being an even number.
So the polynomial must have at least one integer root.
Suppose the polynomial has at least 2 positive roots $x$ and $y$ .
$\frac {1}{y+1} + \frac{1}{2(y+2)} + \frac{1}{3(y+3)} + .... + \frac {1} {n(y+n)} =\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)}  .$
$\Rightarrow (y-x)( \frac {1}{(x+1)(y+1)}+ \frac {1}{2(x+2)(y+2)} +...+\frac {1}{n(x+n)(y+n)} ) = 0 $
$\Rightarrow x=y \Rightarrow$ (no sense)
So for every $n \geq 2$ , equation $( 1 )$ always has a unique positive solution $x_n$
That's all I did , and I have no idea what to do next , I want to establish a relationship between $x_n$ but for $n \geq 3$ the polynomial's solution is a " bad " solution . " so it's very difficult to build. Hope to get help from everyone. Thanks very much .
 A: Let $$f_n(x) = \frac{1}{x+1} + \cdots + \frac{1}{n(x+n)}.$$
Observe that $f_n(x)$ is strictly decreasing in $x$ for $x\ge 0$. Further, for $n\ge 2$, $f_n(0) > 1$ and $\lim_{x\to\infty} f_n(x) = 0$. It follows that there is a unique positive solution $x_n$ to $f_n(x) = 1$.

Let $f$ be defined by
$$f(x) = \sum_{i=1}^\infty \frac{1}{i(x+i)}.$$
Note that $(f_n)$ uniformly converges to $f$ (for $x\ge 0$, $|(f-f_N)(x)| \le \sum_{i=N}^\infty 1/i^2$, which can be made arbitrarily small for sufficiently large $N$).
Let the $x_n$ converge to $l$. Then, uniform convergence implies that $1 = \lim_{n\to\infty} f_n(x_n) = f(l)$. So,
$$
\begin{align*}
1 &= \sum_{i=1}^\infty \frac{1}{i(l+i)} \\
l &= \sum_{i=1}^\infty \frac{1}{i} - \frac{1}{l+i}.
\end{align*}
$$
Equality is seen to hold for $l=1$ since then, the right hand side becomes a telescoping sum which is evaluated to be $1$, and the left is $1$.
Thus, $x_n \to 1$.
A: Since you already find a solution for the first question. I write an answer for the second one. Observe that
$$\sum_{k=1}^n\frac{1}{k\cdot (k+1)}=\sum_{k=1}^n\frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{n+1}.$$ Hence
$$\sum_{k=1}^{\infty}\frac{1}{k\cdot (k+1)}=1.$$
This means for $x=1$ the left handside of your equation tends to 1. One can easily prove, that for $x<1$ and sufficient large $n$ the lefthandside is greater than 1 and for $x>1$ and sufficient large $n$ the lefthandside is less than 1.
A: For you first question, a picture may help clarify what is happening.  This one comes from Wolfram Alpha when $n=7$

In effect your function, let's call it $f_n(x)$, is the sum of $n$ hyperbolae with $n$ vertical asymptotes at $-1,-2,\ldots, -n$, a locally decreasing function of $x$ where it exists,  and with $f_n(x) \to 0^-$ as $x \to - \infty$ and $f_n(x) \to 0^+$ as $x \to + \infty$. So $f_n(x)=1$ gives $n$ real solutions, of which $n-1$ are clearly negative and, since $f_n(0)\ge 1$, one must be non-negative and when $n>1$ is positive.
As others have said the limit as $n\to \infty$ comes when $x=1$ and you get $\sum\limits_{k=1}^\infty \frac{1}{k(k+1)}=1$
