Proof by induction - stuck on simple question! Question:
(Part 1) Show that the inequality 
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(n-1)\cdot n} < \frac{1}{2}
$$ 
works for all natural numbers $n > 2$.
(Part 2) Deduce that for all natural numbers $n$, the following inequality holds:
$$
\frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{n^2} < \frac{7}{4}.
$$
This is a problem I found when rummaging around through old proof by induction questions and has been quite an issue for me. You see, I struggled with the first part, I proved it works for $n=3$ and then wrote out the result for $n=k$ and $n=k+1$ but I couldn't progress from there.
What I did:


*

*When $n = 3$: 
$$
\frac{1}{2\cdot 3} = \frac{1}{6}
$$
and since $1/6 < 1/2$,  the proposition is true for $n=3$.

*Assuming the truth of the proposition when $n = k$ i.e. 
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(k-1)\cdot k} < \frac{1}{2}
$$
and considering $n = k+1$, prove that
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(k+1)\cdot k} < \frac{1}{2}
$$
From there I couldn't seem to find a way to link the assumption into the $n=k+1$ inequality in order to prove that that is also $< 1/2$.


Can't solve the first part in order to link to the second part in order to fulfill the "deduce" part.
Thanks in advance and sorry if it's just a stupid error I've made.

$\require{cancel}$No need for induction here. Just note that\begin{align}\frac1{2\times3}+\frac1{3\times3}+\cdots+\frac1{(n-1)n}&=\frac12-\cancel{\frac13}+\cancel{\frac13}+\cdots-\cancel{\frac1{n-1}}+\cancel{\frac1{n-1}}-\frac1n\\&=\frac12-\frac1n\\&<\frac12.\end{align}And then\begin{align}\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{n^2}&<1+\frac14+\frac1{2\times3}+\cdots+\frac1{n(n-1)}\\&<1+\frac14+\frac12\\&=\frac74.\end{align}
