Question Number 192608 by York12 last updated on 22/May/23
![let k be natural number. defined s_k as the sum of the infinite series s_k =((k^2 −1)/k^0 ) + ((k^2 −1)/k^1 ) + ((k^2 −1)/k^2 ) +... find the value of Σ_(k=1) ^∞ [(s_k /2^(k−1) )] .](https://www.tinkutara.com/question/Q192608.png)
$${let}\:{k}\:{be}\:{natural}\:{number}.\:{defined}\:{s}_{{k}} \:{as}\:{the} \\ $$$${sum}\:{of}\:{the}\:{infinite}\:{series}\:{s}_{{k}} =\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{0}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{1}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\:+… \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[\frac{{s}_{{k}} }{\mathrm{2}^{{k}−\mathrm{1}} }\right]\:\:. \\ $$
Answered by witcher3 last updated on 24/May/23
![S_k =Σ_(n≥0) (((k^2 −1))/k^n ),S_(1=0) ∀k≥2 S_k =(k^2 −1).(1/(1−(1/k)))=k(k+1) Σ_(k≥1) [(S_k /2^(k−1) )]=Σ_(k≥2) [((k(k+1))/2^(k−1) )]=A k(k+1)<2^(k−1) , k=7...proof 6.7<2^6 =64 suppse ∀k≥7 k(k+1)≤2^(k−1) ,(k+1)(k+2)≤2^k we have 2k(k+1)≤2^k 2k^2 +2k=k^2 +3k+2+k^2 −k−2 k^2 −k−2=(k+1)(k−2)≥8.5>0 ⇒true ⇒∀k≥7 ((k(k+1))/2^(k−1) )<1⇒[((k(k+1))/2^ )]=0 A=Σ_(k=1) ^6 [((k(k+1))/2^(k−1) )]+0 =[2]+[(6/2)]+[((12)/4)]+[((20)/8)]+[((30)/(16))]+[((42)/(32))] =12](https://www.tinkutara.com/question/Q192686.png)
$$\mathrm{S}_{\mathrm{k}} =\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{k}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{k}^{\mathrm{n}} },\mathrm{S}_{\mathrm{1}=\mathrm{0}} \\ $$$$\forall\mathrm{k}\geqslant\mathrm{2}\:\mathrm{S}_{\mathrm{k}} =\left(\mathrm{k}^{\mathrm{2}} −\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}}}=\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right) \\ $$$$\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\left[\frac{\mathrm{S}_{\mathrm{k}} }{\mathrm{2}^{\mathrm{k}−\mathrm{1}} }\right]=\underset{\mathrm{k}\geqslant\mathrm{2}} {\sum}\left[\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}^{\mathrm{k}−\mathrm{1}} }\right]=\mathrm{A} \\ $$$$\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)<\mathrm{2}^{\mathrm{k}−\mathrm{1}} , \\ $$$$\mathrm{k}=\mathrm{7}…\mathrm{proof} \\ $$$$\mathrm{6}.\mathrm{7}<\mathrm{2}^{\mathrm{6}} =\mathrm{64} \\ $$$$\mathrm{suppse}\:\forall\mathrm{k}\geqslant\mathrm{7}\:\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\leqslant\mathrm{2}^{\mathrm{k}−\mathrm{1}} ,\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right)\leqslant\mathrm{2}^{\mathrm{k}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{2k}\left(\mathrm{k}+\mathrm{1}\right)\leqslant\mathrm{2}^{\mathrm{k}} \\ $$$$\mathrm{2k}^{\mathrm{2}} +\mathrm{2k}=\mathrm{k}^{\mathrm{2}} +\mathrm{3k}+\mathrm{2}+\boldsymbol{\mathrm{k}}^{\mathrm{2}} −\boldsymbol{\mathrm{k}}−\mathrm{2} \\ $$$$\boldsymbol{\mathrm{k}}^{\mathrm{2}} −\boldsymbol{\mathrm{k}}−\mathrm{2}=\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{k}}−\mathrm{2}\right)\geqslant\mathrm{8}.\mathrm{5}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{true} \\ $$$$\Rightarrow\forall\mathrm{k}\geqslant\mathrm{7}\:\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}^{\mathrm{k}−\mathrm{1}} }<\mathrm{1}\Rightarrow\left[\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}^{} }\right]=\mathrm{0} \\ $$$$\mathrm{A}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\left[\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}^{\mathrm{k}−\mathrm{1}} }\right]+\mathrm{0} \\ $$$$=\left[\mathrm{2}\right]+\left[\frac{\mathrm{6}}{\mathrm{2}}\right]+\left[\frac{\mathrm{12}}{\mathrm{4}}\right]+\left[\frac{\mathrm{20}}{\mathrm{8}}\right]+\left[\frac{\mathrm{30}}{\mathrm{16}}\right]+\left[\frac{\mathrm{42}}{\mathrm{32}}\right] \\ $$$$=\mathrm{12} \\ $$$$ \\ $$
Answered by Rajpurohith last updated on 26/May/23
$${clearly}\:{s}_{{k}} =\left({k}^{\mathrm{2}} −\mathrm{1}\right)\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}^{{i}} }\right)=\left({k}^{\mathrm{2}} −\mathrm{1}\right).\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{k}}}\right) \\ $$$$={k}\left({k}+\mathrm{1}\right) \\ $$$${so}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{s}_{{k}} }{\:\:\mathrm{2}^{{k}−\mathrm{1}} }\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}^{{k}−\mathrm{1}} }\:\:\: \\ $$