Question on an induction

Question

Show that for every $n\ge2$,we have:

\begin{equation} \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{n^2}<1-\frac{1}{n} \end{equation}

Prove by induction:

1. Base case:

\begin{equation} \frac{1}{2^2}<\frac{1}{2}, \end{equation} is valid.

2. We then assume that the statement is also true for $n=k$ \begin{equation} \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{(k)^2}<1-\frac{1}{k} \end{equation}

3. We prove it for $n=k+1$ \begin{equation} \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{(k+1)^2}<1-\frac{1}{k+1} \end{equation}

\begin{equation} \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{(k+1)^2}<\frac{k}{k+1} \end{equation}

\begin{equation} \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots<\frac{k(k+1)-1}{(k+1)^2} \end{equation}

Put $(k+1)=n$ and obtain:

\begin{equation} \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots<\frac{(n-1)(n)-1}{(n)^2} \end{equation}

But I can't solve this one.

Any ideas?

Thanks

Answer 1

When you have an induction don't write the result you want to arrive to and try to get it backward like you did.

Instead start with: let assume it is true for $n=k$ that $S_k<1-\frac 1k$

Then $S_{k+1}=S_k+\frac 1{(k+1)^2}<1-\frac 1k+\frac 1{(k+1)^2}<\cdots<1-\frac 1{k+1}$

so the induction is verified for $n=k+1$ and therefore for all $n$

(cf linked answer by D.W.Farlow for the final detailed calculation)