1
$\begingroup$

I have this problem

Proof that

$\displaystyle\sum_{k=3}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} =48 $

I tryed this:

$\displaystyle\sum_{k=3}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} = \displaystyle\sum_{k=1}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} - \displaystyle\sum_{k=1}^2 k\left({\displaystyle\frac{5}{6}}\right)^{k-3} = \displaystyle\sum_{k=1}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3}-3.48 $, Now

$\displaystyle\sum_{k=1}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} = 1*\left({\displaystyle\frac{5}{6}}\right)^{-2} + 2*\left({\displaystyle\frac{5}{6}}\right)^{-1}+ 3*\left({\displaystyle\frac{5}{6}}\right)^{0} + ......... = \left({\displaystyle\frac{5}{6}}\right)^{-2}(1+2*\left({\displaystyle\frac{5}{6}}\right)^{1}+ 3*\left({\displaystyle\frac{5}{6}}\right)^{2} + .....)$

Let $x=(5/6)$

\begin{eqnarray} 1+2x+3x^2+4x^3+\dots = 1 + x + x^2 + x^3 + \dots \\ + x + x^2+ x^3 + \dots\\ + x^2 + x^3 + \dots \\ +x^3 + \dots \\ + \dots \\ =1 + x + x^2 + x^3+\dots \\ +x(1+x+x^2+\dots) \\ +x^2(1+x+\dots)\\ +x^3(1+\dots)\\ +\dots \\ =(1+x+x^2+x^3+\dots)^2=\frac{1}{(1-x)^2} \end{eqnarray}

then

$\left({\displaystyle\frac{5}{6}}\right)^{-2}\left({\displaystyle\frac{1}{1-5/6}}\right)^{2}= 51.48$ finally

$\displaystyle\sum_{k=3}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} =51.48-3.84 = 48$

Am I right??

Thanks for your help :)

$\endgroup$
2
$\begingroup$

HINT:

If we are allowed to use Calculus,

we know $$\sum_{0\le r<\infty}a t^r=\frac a{1-t} $$ if $|t|<1$

Differentiate both sides wrt $t$ and then multiply by $t^{-4}$


Alternatively, we can use Arithmetico-geometric series formula to show that $$\sum_{0\le r<\infty}rat^r=\frac a{1-r}+\frac r{(1-r)^2}$$

Now divide either sides by $t^4$

$\endgroup$
  • $\begingroup$ On your second last line, you need $S(1-x)$ instead of just $(1-x)$. $\endgroup$ – Andrew D Sep 28 '13 at 18:02
1
$\begingroup$

The crux of your proof -- that

$$ \sum_{n=0}^{+\infty} n x^{n-1} $$

factors as

$$ \sum_{n=0}^{+\infty} (n+1) x^{n} = \left( \sum_{n=0}^{+\infty} x^{n} \right)^2 $$

does indeed look correct.

It might be good to translate your argument into ordinary summation notation:

$$\begin{align} \sum_{n=0}^{+\infty} (n+1) x^{n} &= \sum_{n=0}^{+\infty} \sum_{m=0}^n x^{n} \\&= \sum_{m=0}^{+\infty} \sum_{n=m}^{+\infty} x^n \\&= \sum_{m=0}^{+\infty} x^m \sum_{n=0}^{+\infty} x^n \\&= \left(\sum_{m=0}^{+\infty} x^m\right) \left( \sum_{n=0}^{+\infty} x^n \right) \end{align}$$

Can you see how your argument corresponds to this calculation?

(this sort of calculation doesn't work in general: specifically where I swapped the order of the two sums. But power series, geometric series, and sums of positive numbers are all very nice and allow natural things like this to work)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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