-2

Exercise Calculate $\displaystyle\sum_{n=0}^\infty n \Big( \frac{4}{5}\Big)^{n+1}$

This is what I have so far

$$\displaystyle\sum_{n=0}^\infty n \Big( \frac{4}{5} \Big)^{n+1} = \displaystyle\sum_{n=0}^\infty n \Big( \frac{4}{5} \Big)^n \cdot \Big( \frac{4}{5} \Big)^1 = \frac{4}{5} \displaystyle\sum_{n=0}^\infty n \Big( \frac{4}{5} \Big)^n$$

I know that the series converges to $16$ but I can't quite finish this. This series is so close to being geometric if it weren't for that pesky $n$ attached to it. You could argue that the constants here are $n$ but this $n$ is always changing, so it's not quite what we want. Any advice?

Grigor Hakobyan
  • 2,031
  • 2
    Your tags suggest you are studying power series, particularly Taylor/Maclaurin series. The sum $\sum_{n=1}^\infty nx^{n-1}$ has a nice closed form for $-1 < x < 1$ (think differentiation/integration). – Theo Bendit Nov 10 '21 at 04:55
  • Do you mean to imply that I should consider $\displaystyle\sum_{n=1}^\infty (n-1) \Big( \frac{4}{5} \Big)^n = \displaystyle\sum_{n=1}^\infty (n-1) \Big( \frac{4}{5} \Big)^{n-1} \cdot \Big( \frac{4}{5} \Big)^1 =\frac{4}{5} \displaystyle\sum_{n=1}^\infty (n-1) \Big( \frac{4}{5} \Big)^{n-1}$ ? – Grigor Hakobyan Nov 10 '21 at 04:57
  • 1
    Not quite, but close. Try bringing $\frac{4^2}{5^2}$ out the front instead (and note that the first term of the summation is $0$ and can be ignored). – Theo Bendit Nov 10 '21 at 04:58
  • Also: https://math.stackexchange.com/q/647587/42969, https://math.stackexchange.com/q/30732/42969, https://math.stackexchange.com/q/333192/42969 – Martin R Nov 10 '21 at 06:12

3 Answers3

4

So , $$\sum_{n\geq0}n\left(\frac{4}{5}\right)^{n+1}=1\left(\frac{4}{5}\right)^2+2\left(\frac{4}{5}\right)^3+3\left(\frac{4}{5}\right)^4+.......\infty$$

Now , $$S=1\left(\frac{4}{5}\right)^2+2\left(\frac{4}{5}\right)^3+3\left(\frac{4}{5}\right)^4+..........\infty\tag{1}$$ $$\frac{5S}{4}=1\left(\frac{4}{5}\right)^1+2\left(\frac{4}{5}\right)^2+3\left(\frac{4}{5}\right)^3+.......\infty\tag2$$

Substract equation $(1)$ from $(2)$ . $$\frac{5S}{4}-S=\frac{4}{5}+(2-1)\left(\frac{4}{5}\right)^2+(3-2)\left(\frac{4}{5}\right)^3+..........\infty$$ $$\frac{S}{4}=\frac{4}{5}+\left(\frac{4}{5}\right)^2+\left(\frac{4}{5}\right)^3+.........\infty$$

This is nothing but Infinite GP , where $|r|<1$ as $r=\frac{4}{5}$ [note : sum of infinite GP is $\frac{a}{1-r}$]

So you get $$\frac{S}{4}=\frac{\frac{4}{5}}{1-\frac{4}{5}}\implies S=16$$

RAHUL
  • 1,511
Tryhard
  • 688
3

Expanding the suggestions given in the comment, By the power series expansion we have that, $$\frac{1}{1-x}=\sum_{k\ge 0}x^{k} , |x|<1$$ Differentiation on both sides, $$\frac{1}{(1-x)^{2}}=\sum_{k\ge 0}kx^{k-1}$$ $$\frac{1}{(1-x)^{2}}=\frac{1}{x}\sum_{k\ge 0}kx^{k}$$ Multiply by $x^{2}$ on both sides, $$\frac{x^2}{(1-x)^2}=\sum_{k\ge 0}kx^{k+1}$$ Now according to required question, $x=\frac{4}{5}$. So answer will be, $$\frac{(4/5)^2}{(1-4/5)^2}=\sum_{k\ge 0}k \left(\frac{4}{5}\right)^{k+1}$$ $$\frac{(4/5)^2}{(1-4/5)^2}=\frac{4^2}{1}=16$$

RAHUL
  • 1,511
1

An straight solution was given by @Pi..Mechanic;

Another one is what @TheoBendit try to guide you to that:

$$ (\sum x^n)' = \sum (x^n)' = \sum nx^{(n-1)} $$

Now you can find why @TheoBendit encourage you to:

"Try bringing $\frac{4^2}{5^2}$ out the front instead"

Ali Ashja'
  • 2,786
  • 8
  • 12