# Convergent value of $\sum\limits^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$

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.

• Possible duplicate of Evaluate $\sum_{n=1}^\infty nx^{n-1}$ – David Apr 4 '19 at 1:05
• 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. – Robert Shore Apr 4 '19 at 1:15

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

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.

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.

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