שינויים
/* תרגילים - אי שיוויונים */
*<math>\frac{1^2}{1\cdot 3}+\frac{2^2}{3\cdot 5}+...+\frac{n^2}{(2n-1)(2n+1)}=\frac{n(n+1)}{2(2n+1)}</math>
<math>\frac{1^2}{1\cdot 3}+\frac{2^2}{3\cdot 5}+...+\frac{n^2}{(2n-1)(2n+1)}+\frac{(n+1)^2}{(2n+1)(2n+3)}=\frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{(2n+1)(2n+3)}</math>
<math>=\frac{n(n+1)(2n+3)+2(n+1)^2}{2(2n+1)(2n+3)}=\frac{(n+1)(2n^2+3n+2n+2)}{2(2n+1)(2n+3)}=\frac{(n+1)(n+2)}{2(2n+3}</math>
הראשון לפי הנחת האינדוקציה, השני מכנה משותף, והאחרים פישוט וצמצום
*<math>\Big(1-\frac{1}{(n+1)^2}\Big)\Big(1-\frac{1}{(n+2)^2}\Big)\cdots \Big(1-\frac{1}{(2n)^2}\Big)=\frac{2n+1}{2n+2}</math>
<math>\Big(1-\frac{1}{(n+2)^2}\Big)\Big(1-\frac{1}{(n+3)^2}\Big)\cdots \Big(1-\frac{1}{(2n)^2}\Big)\Big(1-\frac{1}{(2n+1)^2}\Big)\Big(1-\frac{1}{(2n+2)^2}\Big)</math>
<math>=\frac{\Big(1-\frac{1}{(n+1)^2}\Big)\Big(1-\frac{1}{(n+2)^2}\Big)\Big(1-\frac{1}{(n+3)^2}\Big)\cdots \Big(1-\frac{1}{(2n)^2}\Big)}{\Big(1-\frac{1}{(n+1)^2}\Big)}\Big(1-\frac{1}{(2n+1)^2}\Big)\Big(1-\frac{1}{(2n+2)^2}\Big)</math>
<math>=\frac{\frac{2n+1}{2n+2}}{\Big(1-\frac{1}{(n+1)^2}\Big)}\Big(1-\frac{1}{(2n+1)^2}\Big)\Big(1-\frac{1}{(2n+2)^2}\Big)</math>
<math>=\frac{(n+1)^2}{n^2+2n}\cdot\frac{2n+1}{2n+2}\cdot\frac{4n^2+4n}{(2n+1)^2}\cdot\frac{4n^2+8n+3}{(2n+2)^2}=\frac{2n+3}{2n+4}</math>
השוויון הראשון הוא כפל וחלוקה באותו גורם, השני הוא לפי הנחת האינדוקציה, וההמשך זה פישוט וצמצום
==תרגילים - אי שיוויונים==
*<math>\frac{1}{1\cdot 6}+\frac{1}{6\cdot 11} +...+\frac{1}{(5n-4)(5n+1)}<\frac{2n}{5n+1}</math>
<math>\frac{1}{1\cdot 6}+\frac{1}{6\cdot 11} +...+\frac{1}{(5n-4)(5n+1)}+\frac{1}{(5n+1)(5n+6)}<\frac{2n}{5n+1}+\frac{1}{(5n+1)(5n+6)}</math>
<math>=\frac{(2n)(5n+6)+1}{(5n+1)(5n+6)}=\frac{10n^2+12n+1}{(5n+1)(5n+6)}<\frac{10n^2+12n+2}{(5n+1)(5n+6)}=\frac{2n+2}{5n+6}</math>
*<math>\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<\frac{n-1}{n}</math>
<math>\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}+\frac{1}{(n+1)^2}<\frac{n-1}{n}+\frac{1}{(n+1)^2}=\frac{n^3+n^2-1}{n(n+1)^2}<\frac{n^3+n^2}{n(n+1)^2}=\frac{n}{n+1}</math>
*<math>1^2+2^2+...+n^2<\frac{(n+1)^3}{3}</math>
<math>1^2+2^2+...+n^2+(n+1)^2<\frac{(n+1)^3}{3}+(n+1)^2=\frac{n^3+6n^2+9n+4}{3}<\frac{n^3+6n^2+12n+8}{3}=\frac{(n+2)^3}{3}</math>
*<math>\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{13}{24}</math>
<math>\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}>\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}</math>
<math>=\frac{13}{24}+\frac{1}{2n+1}-\frac{1}{2n+2}>\frac{13}{24}</math>
*<math>\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1</math>
<math>\frac{1}{n+2}+...+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}+\frac{1}{3n+4}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}+\frac{1}{3n+4}-\frac{1}{n+1}</math>
<math>>1+\frac{1}{3n+2}+\frac{1}{3n+3}+\frac{1}{3n+4}-\frac{1}{n+1}=1+\frac{2}{3(n+1)(3n+2)(3n+4)}>1</math>
