Is this a proof by induction question? Happened to stumble across this question and to me it immediately made me assume it's a proof my induction question but doesn't seem to be so.

Question :
  Show that for every natural $k$ and $n$ we have
  $$\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}= \frac{n}{k(n+k)}$$
  Hence deduce that the sum above is smaller than $1/k$

I tried to do this by proof by induction but I happened to find no way to link the $n=k$ and  $n+k+1$ part and unless I made a stupid error, I don't see how this can be proved by induction. 
Does anyone know how one would answer this question?
Thanks.
 A: Hint:
You can fix $k \in \mathbb{N}^*$ and show by induction on $n$
$$
\mathcal{P}_n : \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}= \frac{n}{k(n+k)}
$$
To show $\mathcal{P}_n \implies \mathcal{P}_{n+1}$, you have to prove that
$$
\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)} + \frac{1}{(n+k)(n+k+1)}= \frac{n+1}{k(n+1+k)}
$$
using $\mathcal{P}_n$.
A: $$\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)}=\frac{1}{k(k+1)}$$
That's true for all of the terms of the sequence, so
$$\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\frac{1}{(k+2)(k+3)}+...+\frac{1}{(k+n-1)(k+n)} $$
$$=\left(\frac{1}{k}-\frac{1}{k+1}\right)+\left(\frac{1}{k+1}-\frac{1}{k+2}\right)+\left(\frac{1}{k+2}-\frac{1}{k+3}\right)+...+\left(\frac{1}{k+m-1}-
\frac{1}{k+n}\right)$$
$$=\frac{1}{k}-\frac{1}{k+n}=\frac{k+n}{k(k+n)}-\frac{k}{k(k+n)}=\frac{n}{k(k+n)}$$
A: Let $P(n)$ defined as below be true.
$$P(n): \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}= \frac{n}{k(n+k)}$$
Then,
$$P(n+1) : LHS =  \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}+\frac{1}{(n+k)(n+k+1)}$$
$$\implies LHS = \frac{n}{k(n+k)}+\frac{1}{(n+k)(n+k+1)} = \frac{n(n+k+1)+k}{k(n+k)(n+k+1)}=\frac{(n+1)(n+k)}{k(n+k)(n+k+1)}=\frac{n+1}{k(n+k+1)}=RHS$$
So, $P(n)$ is true $\implies P(n+1)$ is true 
