-1
$\begingroup$

This question already has an answer here:

How do you calculate the sum of this series?

$$ \sum_{n=1}^\infty 0.5n \left(\frac{2}{3}\right)^n $$

$\endgroup$

marked as duplicate by Toby Mak, mrtaurho, Javi, Community Apr 30 at 11:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Note that we can pull out the $0.5$ from the summation, and focus on

$$\sum_{n=1}^\infty n \left( \frac 2 3 \right)^n$$

This bears resemblance to a geometric series. We know that, for a geometric series with ratio $r \in (-1,1)$,

$$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$

We want to manipulate this expression into the desired form. First, we see the derivative is given by

$$\sum_{n=0}^\infty nr^{n-1} = \frac{1}{(1-r)^2}$$

and multiplication by $r$ gives

$$\sum_{n=0}^\infty nr^{n} = \frac{r}{(1-r)^2}$$

Since the first summand of the sum ($n=0$) is itself $0$, we can start the summation at $n=1$ if we choose, which would match the form of the sum we choose to focus on. Take $r=2/3$ and we have the "important" sum; once you have the result, multiply by $1/2$ again to account for the pulling out of it at the start, and we're done!

$\endgroup$
  • $\begingroup$ thank you for your answer! My question is, that the derivative(with respect to 'r') of 1/(1-r) is not -1/(1-r)^2? Because (1-r)^(-1) is a composite function. Thank you $\endgroup$ – TTomi Apr 30 at 10:57
  • $\begingroup$ Gotta remember the chain rule, I keep making this exact mistake. XD If we let $$f(r) = \frac{1}{1-r} = (1-r)^{-1}$$ then $$\frac{df}{dr} = -1\cdot (1-r)^{-2} \cdot \frac{d}{dr} (1-r) = (-1) \cdot (1-r)^{-2} \cdot (-1) = (1-r)^{-2}$$ $\endgroup$ – Eevee Trainer Apr 30 at 10:59
  • $\begingroup$ thank you, great, you are totally right!! $\endgroup$ – TTomi Apr 30 at 11:04
1
$\begingroup$

A solution withouth calculus

We want to calculate the sum:

$$\sum_{n=1}^{m} n a^n $$

Notice that:

$$\sum_{n=1}^{m} n a^n=1a^1+2a^2+3a^3+...+m a^m =$$ $$=(a^1+a^2+a^3+...+a^m)+(a^2+a^3+...+a^m)+...+(a^{m-1}+a^m)+(a^m)= a^1(1+a^1+...+a^{m-1})+a^2(1+a^1+..+a^{m-2})+...+a^m(1)$$

By geometric series formula this becomes:

$$a^1 \frac{a^{m}-1}{a-1}+a^2\frac{a^{m-1}-1}{a-1}+...+a^m\frac{a^{1}-1}{a-1}=\sum_{n=1}^{m} a^n \frac{a^{m+1-n}-1}{a-1}$$

Now we're almost done:

$$\sum_{n=1}^{m} a^n \frac{a^{m+1-n}-1}{a-1}=\frac{1}{a-1}\sum_{n=1}^{m} (a^{m+1}-a^{n})=\frac{1}{a-1}\left[\sum_{n=1}^{m} a^{m+1}-\sum_{n=1}^{m}a^{n}\right]=\left[ma^{m+1}-\frac{a^{m+1}-1}{a-1}+1\right]=\left[\frac{ma^{m+2}-ma^{m+1}-a^{m+1}+1+a-1}{(a-1)^2}\right]$$

As $m$ tend to infinity all the exponential terms tend to $0$ since $|a|<1$, so it becomes:

$$\frac{a}{(a-1)^2}$$

If $a=\frac 23$ this is equal to:

$$6$$

Your sum presents also a factor $0.5$ so the final answer is .

$$3$$

:)

$\endgroup$
  • $\begingroup$ wow, great! Thank you Eureka! $\endgroup$ – TTomi Apr 30 at 11:00
0
$\begingroup$

There is a simple trick, for $a < 1$: $$\sum\limits_{n=0}^\infty n a^n = a\partial_a\sum\limits_{n=0}^\infty a^n = a\partial_a \frac{1}{1-a} = \frac{a}{(1-a)^2}$$

$\endgroup$
  • $\begingroup$ thank you denklo! $\endgroup$ – TTomi Apr 30 at 10:57
0
$\begingroup$

Hint:

$$\color{green}{\sum_{n=1}^\infty na^n}=\sum_{n=0}^\infty(n+1)a^{n+1}=a\color{green}{\sum_{n=1}^\infty na^n}+\sum_{n=0}^\infty a^{n+1}.$$

$$|a|<1\implies(1-a)S=\frac a{1-a}.$$

$\endgroup$
  • $\begingroup$ Thank you Yves Daoust! $\endgroup$ – TTomi Apr 30 at 11:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.