# Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$

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.

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

To prove the inequality you were required to prove, note

$$\dfrac1{n+1}+\dfrac1{(n+1)^2}<\dfrac1{n+1}+\dfrac1{(n+1)^2}+\dfrac1{(n+1)^3}+...=\dfrac{\dfrac1{1+n}}{1-\dfrac1{n+1}}=\dfrac1n$$

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 $$n\ge 1$$

• I like this answer but I'm looking for something that is simpler / more obvious. Jun 20, 2019 at 1:40
• 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. 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.

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