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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.

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    See http://math.stackexchange.com/questions/1251544/sum-k-1n-frac1k2-2-frac1n and http://math.stackexchange.com/questions/1220203/proving-1-frac14-frac19-cdots-frac1n2-leq-2-frac1n-for2015-10-25

2 Answers 2

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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.

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    @1337holiday: You are trying to show that $\sum_{k=1}^{n+1}\frac{1}{k^2} \lt 2 - \frac{1}{n+1}.$ You know the first $n$ summands on the left hand side are at most $2 - \frac{1}{n}$, so you know that the *entire* left hand side is at most $\left(2 - \frac{1}{n}\right) + \frac{1}{(n+1)^2}.$ This is **still not** what you want. *Now* what you need to show is that $\left( 2 - \frac{1}{n}\right) + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}.$ You already know that the left hand side of *this* inequality is greater than what you have, but you are trying to show what you have is less than the RHS2011-04-19
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What you write is actually what you are trying to show. However, you are almost there. For the induction step, you get to assume that

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

Then, you need to show the statement holds for $k = n+1$:

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

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    Use the induction hypothesis!2011-04-19