Is it possible to find $a$ and $b$ so that I can get the form $\frac{1}{n+1}=\frac{a}{n}+b$ with constants $a$ and $b$?

Question

Is it possible to find $a$ and $b$ so that I can get the form $\frac{1}{n+1}=\frac{a}{n}+b$ with constants $a$ and $b$?

I tried but failed, because $a$ depends on $n$ in my result: $$\frac{1}{n+1}=\frac{a}{n}+b\\ \Leftrightarrow 1=\frac{a}{n}(n+1)+b(n+1)$$ Set $b=1$: $$1=a+\frac{a}{n}+(n+1)\\ \Leftrightarrow-n^2=an+a=a(n+1)\\ \Leftrightarrow a=\frac{-n^2}{n+1}$$

Answer 1

I think if there exist constants $a,b$ such that:

$$\forall n \in \mathbb{N}, \hspace{0.2cm} \frac{1}{n+1} = \frac{a}{n} + b$$

Then $\forall n \in \mathbb{N}, \hspace{0.1cm} a = n(\frac{1}{n+1} - b)$. Notice that $b\neq 0$ (otherwise $a=\frac{n}{n+1})$. So that by letting $n \to \infty$ $a$ is $+/- \infty$. So the answer is no

Answer 2

It doesn't seem possible...

For $n=0$ you have $\frac 1{n+1}=1$ while $\frac an+b$ is undefined.

For $n\to\infty$ you have $\frac 1{n+1}\to 0$ while $\frac an+b \to b$.