In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.

This is my attempt:

$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\ &=\frac{n+2}{(n+1)^2} \\ &=\frac{n+2}{n^2 + 2n + 1} \\ &<\frac{n+2}{n^2 + 2n} (\text{since } n \geq 1)\\ &=\frac{n+2}{n(n+2)}\\ &=\frac{1}{n} \end{align} $$

I am just wondering if there is a simpler way of doing this.

  • 3
    $\begingroup$ for $n=1$, the inequality $\sum_{k=1}^n \frac{1}{k^2}<2-\frac{1}{n}$ is not true. it only holds for $n>2$. $\endgroup$
    – Kumar
    Jun 20, 2019 at 1:57
  • $\begingroup$ I think the < should be $\geq$. $\endgroup$
    – 1123581321
    Jun 20, 2019 at 2:11
  • $\begingroup$ No, it's correct. Because as $n\rightarrow \infty$ the inequality approaches $\frac{\pi^2}{6}$. $\endgroup$
    – Kumar
    Jun 20, 2019 at 2:27
  • $\begingroup$ Sorry should be $\leq$ $\endgroup$
    – 1123581321
    Jun 20, 2019 at 2:28
  • 2
    $\begingroup$ Nice variety of solutions. I upvoted then all. $\endgroup$ Jun 20, 2019 at 2:56

5 Answers 5


To prove the inequality you were required to prove, note



Instead of comparing $\frac{1}{(n+1)^2} + \frac{1}{n+1}$ and $\frac{1}{n}$, we can compare $\frac{1}{(n+1)^2}$ and $\frac{1}{n}-\frac{1}{n+1}$. Then, what we have is $$\frac{1}{(n+1)^2} = \frac{1}{(n+1)(n+1)} < \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \implies \frac{1}{(n+1)^2} < \frac{1}{n}-\frac{1}{n+1}$$ $$\implies \frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$$

Alternatively, by using the same way, we could try to prove $$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)} > 0$$ For this one, we have $$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)} = \frac{(n+1)^2-n-n(n+1)}{n(n+1)^2} = \frac{1}{n(n+1)^2} > 0$$ since $n \ge 1$. Therefore, the result follows.


Multiply through by $n(n+1)^2$ (a positive number since $n \ge 1$ ) to get $$n+n(n+1) < (n+1)^2$$ $$n < (n+1)^2 - n(n+1)$$ $$n < (n+1)(n+1-n)$$ $$n < n+1$$ Which is true of course for all $n$; in particular $n\ge 1$

  • $\begingroup$ I like this answer but I'm looking for something that is simpler / more obvious. $\endgroup$
    – 1123581321
    Jun 20, 2019 at 1:40
  • $\begingroup$ Another method is to use the common fraction of $n(n+1)^2$ and bring everything to one side: the result is similar: $n(n+1)^2 \gt 0$ which for the given conditions is true. $\endgroup$
    – WaveX
    Jun 20, 2019 at 3:25

Just for the fun of it, you can re-write your inequality as: $$\frac{1}{n}-\frac{1}{n+1}>\frac{1}{(n+1)^2}.$$ This can be re-writen as: $$\int_{n}^{n+1}\frac{1}{x^2}dx>\frac{1}{(n+1)^2},$$ or, if $f(x)=\dfrac{1}{x^2}$: $$\int_{n}^{n+1}f(x)dx>f(n+1),$$ which is obvious if you consider that $f$ is strictly decreasing and positive on $(0,+\infty)$ and that the right-hand side of the inequality is the area of a rectangle with height $f(n+1)$ and a basis of $1$ - see the figure below. enter image description here


Your way is right, but you need to open the sumation from $k=2$: $$\sum_{k=1}^n\frac{1}{k^2}=1+\sum_{k=2}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=2-\frac{1}{n}.$$


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