calculus - Is $(n+1)(h/R)^n$ monotically decreasing when $0< h < R$ and how to prove it?

I have the following alternating series where $R$ is a positive constant:
$$\sum_{n=0}^{\infty}(-1)^n (n+1)\left(\dfrac{h}{R}\right)^n$$

I'm trying to use the "Alternating Series Estimation Theorem" to calculate the sum error at a certain term. To use the theorem on a series $\sum(-1)^{n-1}b_n$ two conditions has to be met first:

1. $b_{n+1} \le b_n\qquad$ (decreasing terms)


2. $\lim_{n\to\infty} b_n = 0$

The second condition can be proved easily by using the ratio test:
\begin{align*}
\lim_{n\to\infty} \left| \dfrac{b_{n+1}}{b_n} \right| &= \lim_{n\to\infty} \left| \dfrac{(n+2)\left(\frac{h}{R}\right)^{n+1}}{(n+1)\left(\frac{h}{R}\right)^n}\right|\\

&= \lim_{n\to\infty} \left(\dfrac{n+2}{n+1}\right) \left|\dfrac{h}{R}\right| \\
&= \dfrac{h}{R}\qquad h > 0
\end{align*}

By the ratio test the series converges absolutely when $\lim_{n\to\infty}|b_{n+1}/b_n| < 1$ therefore the series converges when $h < R$ and:
$$ \lim_{n\to\infty} (-1)^n(n+1)\left(\dfrac{h}{R}\right)^n = 0 \qquad h < R$$

Now $b_n = (n+1)\left(\dfrac{h}{R}\right)^n$ is positive continuous and convergent when $0 < h < R$, but does that means it's also decreasing? if not how do I prove that it's decrerasing?

Answer

You have proved that

$$\lim_{n\to\infty} \left| \dfrac{b_{n+1}}{b_n} \right|=\dfrac{h}{R},$$ with $\dfrac{h}{R}<1 $ then one can see that there exists $N$ such that for all $n\ge N$,
$$
\left| \dfrac{b_{n+1}}{b_n} \right| <1
$$ or
$$
|b_{n+1}|<|b_n|, \quad n\ge N,
$$ thus the given sequence is decreasing (from $N$ onward).