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For all $n\geq 1$, prove with mathematical induction

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$

So far.. I have substituted 1 and saw that the statement is true and I have plugged in n+1 to show that the proof is true for all integers but I don't know how to go about the simplification.. right now I have

LHS: $2-\frac{1}{k}+\frac{1}{(k+1)^2} \leq 2-\frac{1}{k+1}$

Should I try to find common denominators for the left? Step by step explanation please!

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Subtract $2-\frac 1k$ from both sides. –  Mike Miller Jan 30 at 3:34
1  
Please use more descriptive titles in the future. –  mixedmath Jan 30 at 3:41
3  
Also, I see that you seem to be asking your questions from a number theory homework tonight. Do you have an assignment due tomorrow or something? –  mixedmath Jan 30 at 3:42

2 Answers 2

you simply have $$2-\frac{1}{k}+\frac{1}{(k+1)^2} \\ =2-\{\frac{1}{k}-\frac{1}{(k+1)^2}\} \\ =2-\{\frac{(k+1)^2-k}{k(k+1)^2}\}\\ =2-\{\frac{k^2+k+1}{k(k+1)^2}\}\\ \leq 2-\frac{k(k+1)}{k(k+1)^2}$$

Since $$k^2+k+1 \gt k^2+k\\ \frac{k^2+k+1}{k(k+1)^2} \gt \frac{k^2+k}{k(k+1)^2}\\ -\frac{k^2+k+1}{k(k+1)^2} \le -\frac{k^2+k}{k(k+1)^2}$$

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how did you go from this line to the second to last line to the final line. k^2+k+1 doesn't equal k(k+1) –  Lil Jan 30 at 13:11
    
@Lil $k^2+k+1 \gt k^2+k$ –  karakfa Jan 30 at 14:00
    
@karakfa I give him more details –  Semsem Jan 30 at 15:33
    
@Lil I edited it, Do you accept it or you need more details, please feel free to ask me –  Semsem Jan 30 at 15:34

Hint: $\dfrac1{n^2}<\dfrac1{n(n-1)}$ , whose sum is telescopic.

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