What is the sum of the series 1/3 + 2/9 + 3/27 + 4/81 + ........

Question

I remember solving this in highschool , but now I don't remember how to find sum of these kind of series .

I want to find the sum of the general series

Sum $\sum_{n=1}^{\infty} n .a^{-n} = ? $

and $\sum_{n=1}^{N}n. a^{-n} = ? $

Answer 1

Here is an approach that relies on the relationships (i) $k=\sum_{\ell=1}^k(1)$ and (ii) $\sum_{k=\ell}^N r^k=\frac{r^{\ell}-r^{N+1}}{1-r}$ for $|r|<1$. Then, with $r=3^{-1}$ we have

$$\begin{align} \sum_{k=1}^N \frac{k}{3^k}&=\sum_{k=1}^N 3^{-k}\sum_{\ell =1}^k(1)\\\\ &=\sum_{\ell =1}^N \sum_{k=\ell}^N 3^{-k}\\\\ &=\sum_{\ell =1}^N \frac{3^{-\ell}-3^{-(N+1)}}{1-1/3}\\\\ &=\frac32 \sum_{\ell =1}^N \left(3^{-\ell}-3^{-(N+1)}\right)\\\\ &=\frac 32\left(\frac{3^{-1}-3^{-(N+1)}}{1-1/3}\right)-\frac N2 3^{-N}\\\\ &=\frac34 -\frac34 3^{-N}-\frac12 N3^{-N} \end{align}$$

Note as $N\to \infty$ the sum of interest approaches $3/4$.

Answer 2

You need to know off by heart $$\frac{1}{1-x}=1+x+x^2+x^3+\dots$$ a formula that is constantly coming up in all areas of maths. You can easily prove it by comparing the sum $s$ with the sum $x\cdot s$.

Ideally you would also remember $$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3\dots$$ which is also extremely useful. Since the tags suggest you have some calculus, the easiest way to get it is to differentiate the first series. Note that both series are absolutely convergent for $|x|<1$. Alternatively you can get it by taking the sum as $s$ and some algebra like that suggested for the first series (that hint should be enough given that $(1-x)^2=1-2x+x^2$).

Answer 3

You need to solve the second sum first.

It runs this way: you learnt in high school the factorisation formula $$1-x^n=(1-x)(1+x+\dots+x^{n-1}),$$ which can be rewritten as $$\frac1{1-x}=1+x+\dots+x^{n-1}+\frac{x^n}{1-x}.$$ From this you deduce that the sum $\;\displaystyle\sum_{i=0}^n x^i$ has a limit (‘the series converges’) if and only if $\lvert x\rvert<1$, and the limit is $\dfrac1{1-x}$.

As to the sum $\;\displaystyle\sum_{i=\color{red}1}^n x^i$, just factor $x$ to obtain it converges to $\dfrac x{1-x}$ under the same assumptions.

For the actual problem: Your series has the form $\displaystyle\sum_{n\ge 1}nx^n$. Factor $x$ $$\sum_{n\ge 1}nx^n=x\sum_{n\ge 1}nx^{n-1}=x\sum_{n\ge 0}(x^{n})'=x\Bigl(\sum_{n\ge 0}x^{n}\Bigr)'=x\Bigl(\frac1{1-x}\Bigr)'=\frac x{(1-x)^2}.$$

Answer 4

Here's an approach that requires the value of the geometric series $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ for $|x|<1$, and termwise differentiation of power series.

To compute the value of \begin{align} S&= \sum_{n=1}^\infty n\left(\frac{1}{3}\right)^n, \end{align} define, for $|x|<1$ $$f(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ Then note that, for $|x|<1$, we have $$xf'(x) = x\sum_{n =1}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}$$ Evaluating this at $x=1/3$ gives $$\frac{1}{3}f'(1/3) = \frac{\frac{1}{3}}{\left(1-\frac{1}{3}\right)^2} = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{3}{4}$$

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To compute the sum $\sum_{n=1}^\infty n a^{-n}$, use the approach above, and we have (upon substitution of $a^{-1}$ for $x$) $$\sum_{n=1}^\infty n a^{-n} = a^{-1}f'(a^{-1}) = \frac{a^{-1}}{\left(1-a^{-1}\right)^2}$$ for $|a| > 1$ (to guarantee convergence, since this is equivalent to $\left|a^{-1}\right|<1$).

Answer 5

Let $$S_m=\sum_{k=m}^\infty kr^k=\sum_{k=m-1}^\infty (k+1)r^{k+1}=mr^m+r\sum_{k=m}^\infty (k+1)r^k.$$

Then

$$S_m-mr^m-rS_m=r\sum_{k=m}^\infty r^k=\frac{r^{m+1}}{1-r}$$ and

$$S_m=\frac{r^m(m-(m-1)r)}{(1-r)^2}.$$

So

$$\sum_{k=1}^\infty kr^k=S_1=\frac r{(1-r)^2}$$ and

$$\sum_{k=1}^n kr^k=S_1-S_{n+1}=\frac{r-r^{n+1}(n+1-nr)}{(1-r)^2}.$$