1
$\begingroup$

The sum in question is:

$$\sum^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$$

It passes the ratio test:

\begin{align} &\lim_{n\rightarrow \infty}\frac{(n+1)\left(\frac{5}{6}\right)^{n}}{n\left(\frac{5}{6}\right)^{n-1}}\\ =\frac{5}{6}&\lim_{n\rightarrow \infty}\frac{(n+1)}{n}\frac{\left(\frac{5}{6}\right)^{n}}{\left(\frac{5}{6}\right)^{n}}\\ =\frac{5}{6}&\lim_{n\rightarrow \infty}(1+ \frac{1}{n})\\ =\frac{5}{6} &< 1\Rightarrow \text{convergent} \end{align} But now I do not know how to find the convergent value.

$\endgroup$
2
  • 3
    $\begingroup$ Possible duplicate of Evaluate $\sum_{n=1}^\infty nx^{n-1}$ $\endgroup$ – David Apr 4 '19 at 1:05
  • $\begingroup$ I don't think it's a duplicate. The other question asks how to evaluate the sum without relying on the derivative of the closed form. $\endgroup$ – Robert Shore Apr 4 '19 at 1:15
4
$\begingroup$

$$f(x)= \frac{1}{1-x}=\sum_{n=0}^\infty x^n\\ f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\\ f'(5/6)=\sum_{n=1}^\infty n(\frac{5}{6})^{n-1}= 36.$$

$\endgroup$
2
$\begingroup$

Hmm so write $f(x) = \sum_{n=1}^{\infty}x^n$. We know how to write this as a geometric series sum, and know the answer is $\frac{x}{1-x}$ (whenever $|x|<1$, which is the case with $x=5/6$ fortunately.

Now we will differentiate $f(x)$ and set $x=5/6$. We are allowed to do this by Taylor's theorem and term by term differentiation, because our series is just a Taylor series for $f(x)$.

If we differentiate the expression $\frac{x}{1-x}$ in $x$ and set $x=5/6$ we will get the answer.

$\endgroup$
2
$\begingroup$

Recall the geometric sum:

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

Take the derivative of both sides:

$$\sum_{k=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$$

Your sum appears when $x=5/6$. The above sums only hold if $|x|<1$ which does hold here.

$\endgroup$
1
  • $\begingroup$ Your answer was just as good as Robert Shore's but I marked his as correct, thus is life. $\endgroup$ – Tsangares Apr 4 '19 at 1:31

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.