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Let ${\displaystyle S_{n}=\sum\limits_{k=0}^{n}\dfrac{1}{3^k}}$ and ${\displaystyle S'_{n}=\sum\limits_{k=0}^{n}\dfrac{k}{3^k}}$

  • Show that $(S_{n})_{n}$ is convergent and calculate its limit
  • Study $(S'_n)_{n}$

The original of text enter image description here

for $S_n$ is Geometric series then $S_n=\sum_{k=0}^{n}\dfrac{1}{3^k}=\sum_{k=0}^{n}\left(\dfrac{1}{3}\right)^{k}=\left(\dfrac{1}{3}\right)^{0}\times \dfrac{1-\left(\dfrac{1}{3}\right)^{n+1}}{1-\left(\dfrac{1}{3}\right)}=\dfrac{3}{2}( 1-\left(\dfrac{1}{3}\right)^{n+1} )$

when $n\to +\infty\quad S_n \to \dfrac{3}{2}$

for second question here is another way:

\begin{align*} S'_n+S_n&=\sum_{k=0}^{n} \frac{k+1}{3^k}\\ S'_n&=\sum_{k=0}^{n} \frac{k+1}{3^k}-S_n\\ S'_n&=3\sum_{k=0}^{n} \frac{k+1}{3^{k+1}}- S_n\\ &=3\sum_{k=1}^{n+1} \frac{k}{3^{k}}- S_n\\ &=3S'_n+\frac{n+1}{3^n}-S_n \end{align*}

(note that $S'_n=\sum_{k=0}^{n}\frac{k}{3^k}=\sum_{k=1}^{n} \frac {k}{3^k})$

$$S'_n=\frac{1}{2} ( S_n-\frac{n+1}{3^n})$$ Thus $S'_n$ is converge to $\dfrac{3}{4}$

any help would be appreciated

Educ
  • 4,780

2 Answers2

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Your answer is correct for the first series. For the second let

$$u(x)=\sum_{k=0}^\infty x^k=\frac1{1-x},\quad |x|<1$$ then

$$u'(x)=\sum_{k=0}^\infty k x^{k-1}=\frac1{(1-x)^2}$$ so for $x=\frac13$ we get

$$\lim_{n\to\infty}S'_n=\frac{\frac13}{(1-\frac13)^2}=\frac34$$

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Hint: $\dfrac{k}{3^k} = k\left(\dfrac{1}{3}\right)^k$, and

$\displaystyle \sum_{k=1}^n kx^k = x\left(1+x+x^2+\cdots x^n\right)'$

DeepSea
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  • how did you get $\displaystyle \sum_{k=1}^n kx^k = x\left(1+x+x^2+\cdots x^n\right)'$ can you elaborate this – Educ Dec 14 '14 at 09:27
  • If you take derivative term by term and times the result by $x$ to see what you get? – DeepSea Dec 14 '14 at 22:32