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1-2-3-4-7-8-15-16-




Question Number 145444 by mathdanisur last updated on 04/Jul/21
(1/2) - (3/4) + (7/8) - ((15)/(16)) + ... = ?
$$\frac{\mathrm{1}}{\mathrm{2}}\:-\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{7}}{\mathrm{8}}\:-\:\frac{\mathrm{15}}{\mathrm{16}}\:+\:…\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 05/Jul/21
S =  Σ_(n=0) ^∞ (−1)^n ((2^(n+1) −1)/2^(n+1) )  S =  Σ_(n=0) ^∞ (−1)^n (1−(1/2^(n+1) ))  S =  Σ_(n=0) ^∞ (−1)^n +Σ_(n=1) ^∞ (((−1)^n )/2^n )  Σ_(n=0) ^∞ (−1)^n  diverges  Σ_(n=1) ^∞ (((−1)^n )/2^n ) = −1+Σ_(n=0) ^∞ (((−1)^n )/2^n )  = −1+(1/(1+(1/2))) = −(1/3) converges  S diverges
$$\mathrm{S}\:=\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$$\mathrm{S}\:=\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\right) \\ $$$$\mathrm{S}\:=\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\mathrm{diverges} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\:=\:−\mathrm{1}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} } \\ $$$$=\:−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{converges} \\ $$$$\mathrm{S}\:\mathrm{diverges} \\ $$

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