# Proof by induction of $\sum_{n=1}^{\infty }\left ( x+1 \right )^{-n}=\frac{1}{x}$

I've been trying to work on some proof for class and I basically want to prove that:

$$\sum_{n=1}^{\infty }\left ( x+1 \right )^{-n}=\frac{1}{x},\quad \text{where} \quad x \in \mathbb{Z^{+}}.$$

So far I've been trying to use proof by induction, but I can't seem to get anywhere as it has no final term. Does anyone have any idea how I could go about proving this?

Guide:

Use geometric series.

Notice that if $x \in \mathbb{Z}^+$, then $0<\frac{1}{x+1} < 1$.

In general, if $a \in \mathbb{C}$ with $\vert a \vert < 1$ and $N \in \mathbb{N}^{\ast}$:

\begin{align*} (1-a) \sum_{k=0}^{N} a^k & = \sum_{k=0}^{N} a^k - \sum_{k=0}^{N} a^{k+1} \\[2mm] & = \sum_{k=0}^{N} a^k - \sum_{k=1}^{N+1} a^k \\[2mm] & = 1 - a^{N+1}. \end{align*}

Because $\vert a \vert < 1$, $a^{N+1} \to 0$ as $N \to +\infty$.

This proves that:

$$\sum_{k=0}^{+\infty} a^k = \frac{1}{1-a}.$$

Apply this result with $a = 1/(1+x)$, $x \in \mathbb{Z}^{+}$.

It's obvious that our series converges.

Now, let $\sum\limits_{n=1}^{+\infty}\frac{1}{(1+x)^n}=A$.

Thus, $$(1+x)A=1+A$$