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

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