Question Number 197349 by SANOGO last updated on 14/Sep/23
$${calcul}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+{oo}} {\sum}}\left(−\mathrm{1}\right)^{{n}\:} \frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$
Answered by MM42 last updated on 14/Sep/23
$$\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$${s}_{{n}} =−\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)−… \\ $$$${s}_{{n}} =−\mathrm{1}\pm\frac{\mathrm{1}}{{n}+\mathrm{1}}\Rightarrow{s}={lim}_{{n}\rightarrow\infty} {s}_{{n}} =−\mathrm{1} \\ $$$$ \\ $$
Commented by SANOGO last updated on 14/Sep/23
$${merci} \\ $$
Commented by MathematicalUser2357 last updated on 16/Sep/23
$${ronaldo} \\ $$