Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 81733 by naka3546 last updated on 15/Feb/20

Σ_(n=1) ^∞ (n/((n+1)∙5^n )) = ?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}}{\left({n}+\mathrm{1}\right)\centerdot\mathrm{5}^{{n}} }\:\:=\:\:? \\ $$

Commented by mr W last updated on 15/Feb/20

(1/(1−x))=1+x+x^2 +x^3 +...=Σ_(n=0) ^∞ x^n (1/((1−x)^2 ))=Σ_(n=1) ^∞ nx^(n−1) (x/((1−x)^2 ))=Σ_(n=1) ^∞ nx^n ∫_0 ^x (x/((1−x)^2 ))dx=Σ_(n=1) ^∞ n∫_0 ^x x^n dx [ln (1−x)+(1/(1−x))]_0 ^x =Σ_(n=1) ^∞ ((nx^(n+1) )/(n+1)) ln (1−x)+(x/(1−x))=Σ_(n=1) ^∞ ((nx^(n+1) )/(n+1)) ((ln (1−x))/x)+(1/(1−x))=Σ_(n=1) ^∞ ((nx^n )/(n+1)) x=(1/5) 5(ln (4/5)+(1/4))=Σ_(n=1) ^∞ (n/((n+1)5^n ))

$$\frac{\mathrm{1}}{\mathrm{1}−{x}}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}−\mathrm{1}} \\ $$$$\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}} \\ $$$$\int_{\mathrm{0}} ^{{x}} \frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}\int_{\mathrm{0}} ^{{x}} {x}^{{n}} {dx} \\ $$$$\left[\mathrm{ln}\:\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right]_{\mathrm{0}} ^{{x}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−{x}\right)+\frac{{x}}{\mathrm{1}−{x}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{ln}\:\left(\mathrm{1}−{x}\right)}{{x}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}} }{{n}+\mathrm{1}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{5}\left(\mathrm{ln}\:\frac{\mathrm{4}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\left({n}+\mathrm{1}\right)\mathrm{5}^{{n}} } \\ $$

Commented by mathmax by abdo last updated on 15/Feb/20

S =Σ_(n=1) ^∞ ((n+1−1)/((n+1)5^n )) =Σ_(n=1) ^∞ (1/5^n )−Σ_(n=1) ^∞ (1/((n+1)5^n )) Σ_(n=1) ^∞ (1/5^n ) =Σ_(n=0) ^∞ ((1/5))^n −1 =(1/(1−(1/5)))−1 =(5/4)−1 =(1/4) Σ_(n=1) ^∞ (1/((n+1)5^n )) =Σ_(n=2) ^∞ (1/(n5^(n−1) )) =5 Σ_(n=2) ^∞ (1/n) ((1/5))^n =5{ Σ_(n=1) ^∞ (1/n)((1/5))^n −(1/5)} let find w(x)=Σ_(n=1) ^∞ (x^n /n) {with ∣x∣<1} w^′ (x)=Σ_(n=1) ^∞ x^(n−1) =Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒w(x)=−ln(1−x)+c c=w(0)=0 ⇒w(x)=−ln(1−x)⇒Σ_(n=1) ^∞ (1/((n+1)5^n )) =5(−ln(1−(1/5)))−1 =−5ln((4/5))−1 ⇒ S=(1/4) +5ln((4/5))+1 =(5/4) +5ln((4/5))

$${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}+\mathrm{1}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\mathrm{5}^{{n}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{5}^{{n}} }−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\mathrm{5}^{{n}} } \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{5}^{{n}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}} −\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}}−\mathrm{1}\:=\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\mathrm{5}^{{n}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}\mathrm{5}^{{n}−\mathrm{1}} }\:=\mathrm{5}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\:\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\mathrm{5}\left\{\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}} −\frac{\mathrm{1}}{\mathrm{5}}\right\}\:\:{let}\:{find}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\left\{{with}\:\mid{x}\mid<\mathrm{1}\right\} \\ $$$${w}^{'} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}−\mathrm{1}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{w}\left({x}\right)=−{ln}\left(\mathrm{1}−{x}\right)+{c} \\ $$$${c}={w}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{w}\left({x}\right)=−{ln}\left(\mathrm{1}−{x}\right)\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\mathrm{5}^{{n}} } \\ $$$$=\mathrm{5}\left(−{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)\right)−\mathrm{1}\:=−\mathrm{5}{ln}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)−\mathrm{1}\:\Rightarrow\:{S}=\frac{\mathrm{1}}{\mathrm{4}}\:+\mathrm{5}{ln}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)+\mathrm{1} \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}\:+\mathrm{5}{ln}\left(\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com