summation - Proof by Induction for inequality, $sum_{k=1}^nk^{-2}lt2-(1/n)$

Let $n$ be a positive natural number, $n\ge 2$. Then $\displaystyle\sum_{k=1}^n \frac{1}{k^2} \lt 2 - \frac{1}{n}.$

The basis step was easy but could someone give me a hint in the right direction as to how to do the induction step?

I tried this:

$\displaystyle\sum_{k=1}^k \frac{1}{k^2} + \frac{1}{(k + 1)^2} \lt 2 - \frac{1}{k + 1}$

But it's getting me nowhere or I am doing something wrong. I am no expert so a clear explanation would be appreciated. Thanks.

Answer

Here is how the induction step should look:

$$ \text{Assume } \sum_{k = 1}^{n} \frac{1}{k^2} < 2 - \frac{1}{n}.$$

Then,

$$ \sum_{k = 1}^{n+1} \frac{1}{k^2} = \sum_{k = 1}^{n} \frac{1}{k^2} + \frac{1}{(n+1)^2} < 2 - \frac{1}{n} + \frac{1}{(n+1)^2}$$

Now the problem is reduced to showing that

$$ - \frac{1}{n} + \frac{1}{(n+1)^2} \leq - \frac{1}{n+1} $$

which is easy to show with some algebra. The point is that you have to use the assumption that it works for $n$. Also, when you use $k$ as an index over which you are summing, you should not use $k$ anywhere else like you did above.

Hope this helps.