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Question Number 133231 by SOMEDAVONG last updated on 20/Feb/21

Give S_n =((n+1)/2^(n+1) )Σ_(k=1) ^n (2^k /k)  .Find lim_(n→+∝) S_n  .

$$\mathrm{Give}\:\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{2}^{\mathrm{k}} }{\mathrm{k}}\:\:.\mathrm{Find}\:\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}S}_{\mathrm{n}} \:. \\ $$

Answered by Ar Brandon last updated on 21/Feb/21

Σ_(k=1) ^n x^(k−1) =((x^n −1)/(x−1))  Σ_(k=1) ^n (x^k /k)=∫ ((x^n −1)/(x−1))dx  Σ_(k=1) ^n (2^k /k)=∫ ((x^n −1)/(x−1))dx_(x=2)

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{x}^{\mathrm{k}−\mathrm{1}} =\frac{\mathrm{x}^{\mathrm{n}} −\mathrm{1}}{\mathrm{x}−\mathrm{1}} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{x}^{\mathrm{k}} }{\mathrm{k}}=\int\:\frac{\mathrm{x}^{\mathrm{n}} −\mathrm{1}}{\mathrm{x}−\mathrm{1}}\mathrm{dx} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{2}^{\mathrm{k}} }{\mathrm{k}}=\int\:\frac{\mathrm{x}^{\mathrm{n}} −\mathrm{1}}{\mathrm{x}−\mathrm{1}}\mathrm{dx}_{\mathrm{x}=\mathrm{2}} \\ $$

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