Σ_(n=1) ^∞ (n/((n+1)∙5^n )) = ?


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

$\underset{n = 1}{\sum^{\infty}} \frac{n}{(n + 1) \cdot 5^{n}} = ?$
Commented by mr W last updated on 15/Feb/20

$\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + . . . = \underset{n = 0}{\sum^{\infty}} x^{n}$ $\frac{1}{{(1 - x)}^{2}} = \underset{n = 1}{\sum^{\infty}} {n x}^{n - 1}$ $\frac{x}{{(1 - x)}^{2}} = \underset{n = 1}{\sum^{\infty}} {n x}^{n}$ $\int_{0}^{x} \frac{x}{{(1 - x)}^{2}} d x = \underset{n = 1}{\sum^{\infty}} n \int_{0}^{x} x^{n} d x$ ${[\ln (1 - x) + \frac{1}{1 - x}]}_{0}^{x} = \underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n + 1}}{n + 1}$ $\ln (1 - x) + \frac{x}{1 - x} = \underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n + 1}}{n + 1}$ $\frac{\ln (1 - x)}{x} + \frac{1}{1 - x} = \underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n}}{n + 1}$ $x = \frac{1}{5}$ $5 (\ln \frac{4}{5} + \frac{1}{4}) = \underset{n = 1}{\sum^{\infty}} \frac{n}{(n + 1) 5^{n}}$
Commented by mathmax by abdo last updated on 15/Feb/20
$S =Σ_(n=1) ^∞ ((n+1−1)/((n+1)5^n )) =Σ_(n=1) ^∞ (1/5^n )−Σ_(n=1) ^∞ (1/((n+1)5^n )) Σ_(n=1) ^∞ (1/5^n ) =Σ_(n=0) ^∞ ((1/5))^n −1 =(1/(1−(1/5)))−1 =(5/4)−1 =(1/4) Σ_(n=1) ^∞ (1/((n+1)5^n )) =Σ_(n=2) ^∞ (1/(n5^(n−1) )) =5 Σ_(n=2) ^∞ (1/n) ((1/5))^n =5{ Σ_(n=1) ^∞ (1/n)((1/5))^n −(1/5)} let find w(x)=Σ_(n=1) ^∞ (x^n /n) {with ∣x∣<1} w^′ (x)=Σ_(n=1) ^∞ x^(n−1) =Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒w(x)=−ln(1−x)+c c=w(0)=0 ⇒w(x)=−ln(1−x)⇒Σ_(n=1) ^∞ (1/((n+1)5^n )) =5(−ln(1−(1/5)))−1 =−5ln((4/5))−1 ⇒ S=(1/4) +5ln((4/5))+1 =(5/4) +5ln((4/5))$
$S = \sum_{n = 1}^{\infty} \frac{n + 1 - 1}{(n + 1) 5^{n}} = \sum_{n = 1}^{\infty} \frac{1}{5^{n}} - \sum_{n = 1}^{\infty} \frac{1}{(n + 1) 5^{n}}$ $\sum_{n = 1}^{\infty} \frac{1}{5^{n}} = \sum_{n = 0}^{\infty} {(\frac{1}{5})}^{n} - 1 = \frac{1}{1 - \frac{1}{5}} - 1 = \frac{5}{4} - 1 = \frac{1}{4}$ $\sum_{n = 1}^{\infty} \frac{1}{(n + 1) 5^{n}} = \sum_{n = 2}^{\infty} \frac{1}{n 5^{n - 1}} = 5 \sum_{n = 2}^{\infty} \frac{1}{n} {(\frac{1}{5})}^{n}$ $= 5 {\sum_{n = 1}^{\infty} \frac{1}{n} {(\frac{1}{5})}^{n} - \frac{1}{5}} l e t f i n d w (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n} {w i t h ∣ x ∣< 1}$ $w^{^{'}} (x) = \sum_{n = 1}^{\infty} x^{n - 1} = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow w (x) = - l n (1 - x) + c$ $c = w (0) = 0 \Rightarrow w (x) = - l n (1 - x) \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{(n + 1) 5^{n}}$ $= 5 (- l n (1 - \frac{1}{5})) - 1 = - 5 l n (\frac{4}{5}) - 1 \Rightarrow S = \frac{1}{4} + 5 l n (\frac{4}{5}) + 1$ $= \frac{5}{4} + 5 l n (\frac{4}{5})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum