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Prove by mathematical induction. I was hoping if someone could give me a hint on how to solve this problem.

$$\frac{1}{1^2}+ \frac {1}{2^2} + ....+ \frac{1}{n^2} < 2 - \frac {1}{n} $$ for all integers $n\geq 2$.

Brian Fitzpatrick
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user124557
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  • The inductive hypothesis would look something like "Assume that $1/1^2 + 1/2^2 + \cdots + 1/(n-1)^2 < 2 - 1/(n-1)$." If you add $1/n^2$ to both sides of this inequality, you get what you want on the left-hand side - can you compare the right-hand side to your goal? – ajd Feb 12 '14 at 02:45
  • So adding $1/n^2$ would allow me to go through the k+1 process? – user124557 Feb 12 '14 at 02:50
  • Well, with induction, you get to assume the statement for the $n-1$ case and then you have to prove it for the $n$ case (or, equivalently, assume it for the $n$ case and then prove it for the $n+1$ case). One way to do that is to take the statement for the $n-1$ case and then derive the statement for the $n$ case. Adding $1/n^2$ to both sides is the first step for deriving the $n$ case from the $n-1$ case. – ajd Feb 12 '14 at 02:54

1 Answers1

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  1. check this for $k=2$
  2. suppose it is true for $n=k$ i.e. $$\frac{1}{1^2}+ \frac {1}{2^2} + ....+ \frac{1}{k^2} < 2 - \frac {1}{k}$$
  3. Prove it for $n=k+1$ using 2

$$ \implies \frac{1}{1^2}+ \frac {1}{2^2} + ....+ \frac{1}{k^2}+\frac{1}{(k+1)^2} < 2 - \frac {1}{k}+\frac{1}{(k+1)^2} \\ =2 -( \frac {1}{k}-\frac{1}{(k+1)^2}) \\=2 -\frac{k^2+k+1}{k(k+1)^2} \\ < 2 -\frac{k^2+k}{k(k+1)^2} \\= 2 -\frac{1}{k+1}$$

Semsem
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