Question Number 220264 by Tawa11 last updated on 10/May/25
Answered by SdC355 last updated on 10/May/25
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{h}+\mathrm{9}\right)\left(\mathrm{2}{h}+\mathrm{11}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}{h}+\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{2}{h}+\mathrm{11}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{11}}−\frac{\mathrm{1}}{\mathrm{13}}+\frac{\mathrm{1}}{\mathrm{13}}−\frac{\mathrm{1}}{\mathrm{15}}+….\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{22}} \\ $$
Commented by Tawa11 last updated on 10/May/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$