# Thread: Induction proof concerning a finite sum inequality

1. ## Induction proof concerning a finite sum inequality

Could someone help me with the style? Specifically, I don't like (4) and I wish there was a way to express (1) through a series of transformation on (3). Keeping the number of steps about the same or less of course.

2. here was the same post below. Sorry for the inconvenience

3. this is how I got it,

assume (1) is true up to n=p

then,
$\sum_{i=0}^{p} \frac{1}{i^2} \leq 2-\frac{1}{p}$

adjust and subtract both sides by $\frac{1}{(p+1)}$
$\sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)}\leq 2-\frac{1}{(p+1)}$

since (8)
$\sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)}\geq \frac{1}{(p+1)^2}+\sum_{i=0}^{p} \frac{1}{i^2}$
$\sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)}\geq \sum_{i=0}^{p+1} \frac{1}{i^2}$
$\sum_{i=0}^{p+1} \frac{1}{i^2} \leq \sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)} \leq 2-\frac{1}{(p+1)}$

$\sum_{i=0}^{p+1} \frac{1}{i^2} \leq 2-\frac{1}{(p+1)}$ which prove correct for n=p+1

4. Try to mimick the style of typical proofs by induction:

And in general, by mimicking the style of others one can eventually develop one's own style, or so I believe.

5. A better bound:

$\sum_{i=0}^{p}\frac{1}{i^2}=1+\frac{1}{2^2}+\frac{ 1}{3^2}+\frac{1}{4^2}+...+\frac{1}{p^2}<1+\frac{1} {2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^p}= 2-(\frac{1}{2})^p$

6. Originally Posted by Ontolog

Could someone help me with the style? Specifically, I don't like (4) and I wish there was a way to express (1) through a series of transformation on (3). Keeping the number of steps about the same or less of course.
Alternatively,

P(k)

$\sum_{i=1}^{k}\frac{1}{i^2}\le\ 2-\frac{1}{k}$

P(k+1)

$\sum_{i=1}^{k+1}\frac{1}{i^2}\le\ 2-\frac{1}{k+1}$

Try to show that IF P(k) is true, THEN P(k+1) will also be true.

If P(k) is true, then P(k+1) is

$x+\frac{1}{(k+1)^2}\le\ 2-\frac{1}{k+1}\;\;?$

where $x\le\ 2-\frac{1}{k}$

$x\le\ 2-\frac{1}{k+1}-\frac{1}{(k+1)^2}\;\;?\;\;\Rightarrow\ x\le\ 2-\left[\frac{k+1+1}{(k+1)^2}\right]\;\;?$

$x\le\ 2-\left[\frac{1}{\left(\frac{[k+1]^2}{k+2}\right)}\right]\;\;?\;\;\Rightarrow\ x\le\ 2-\frac{1}{\left(\frac{k^2+2k+1}{k+2}\right)}\;\;?$

$\Rightarrow\ x\le\ 2-\frac{1}{\left(\frac{k[k+2]+1}{k+2}\right)}\;\;?$

True, since

$\frac{k(k+2)+1}{k+2}>\frac{k(k+2)}{k+2}$

so the above line is subtracting a smaller value from 2 than 1/k

7. The inequality $\sum_{i=1}^{n}\frac{1}{i^2}<2-\frac{1}{n}$ can be established without using induction.

From $i^2>(i-1)i$, it follows $\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}<1 +\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{ 1}{(n-1)\cdot n}$, which is a telescoping sum.(a)
Therefore, $\sum_{i=1}^{n}\frac{1}{i^2}<1+1-\frac{1}{n}=2-\frac{1}{n}$. (To Also sprach Zarathustra, $2-\frac{1}{n}<2-\frac{1}{2^n}$ for $n\geq1$)

(a) $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1 }{(n-1)\cdot n}=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots+(\frac{1}{n-1}-\frac{1}{n})$; the terms, except the first and last, cancel.