Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
Subtract $2-\frac 1k$ from both sides. –  Mike Miller Jan 30 at 3:34
Please use more descriptive titles in the future. –  mixedmath Jan 30 at 3:41
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 $$k^2+k+1 \gt k^2+k\\ \frac{k^2+k+1}{k(k+1)^2} \gt \frac{k^2+k}{k(k+1)^2}\\ -\frac{k^2+k+1}{k(k+1)^2} \le -\frac{k^2+k}{k(k+1)^2}$$

share|improve this answer
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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.