Proof that $\sum \limits_{k=0}^{\infty} \left( k+1\right) \cdot \left( -x\right)^{k}$ converges I'm asked to proof the convergence of $\sum \limits_{k=0}^{\infty} \left( k+1\right) \cdot \left( -x\right)^{k}$ to $0$ for $x\in\left( 0,1 \right)$
Well, I've started with the alternating series test:
$$
\sum_{k=0}^{\infty} \left( k+1\right) \cdot \left( -x\right)^{k}
 = \sum_{k=0}^{\infty} \left( k+1\right) \cdot x^{k}
                                         \cdot \left( -1\right)^{k}
$$
because it turns out for $x\in\left( 0,1 \right)$ the sequence $\left( a_{k} \right)_{k\in\mathbb{N}}=\left( k+1\right)\cdot x^{k}$:
$(1)$ decreases monotonically $\forall k\geq k_{0}(x)$ and
$(2)$ $\lim\limits_{k\to\infty} a_{k} = 0$.
If I look at $\left( a_{k} \right)_{k\in\mathbb{N}}$ as $f_{k}(x)=\left( k+1\right)\cdot x^{k}$, the derivative for k is given by $g(x)=x^{k}\left( \ln(x)k+\ln(x)+1\right)$. Because of $x^{k}\left( \ln(x)k+\ln(x)+1\right)=0 \Leftrightarrow k=0 \text{ or } k=-\frac{\ln(x)+1}{ln(x)}$ and $g(x)\leq 0$ for $k\geq-\frac{\ln(x)+1}{ln(x)}$, I can conclude $(1)$.
I think I'm missing the forest through the trees, but how to proof $(2)$? I've tried to use $\varepsilon$-criterion, but unfortunately I was not successful. If someone could give me a hint, I would be very thankful.
 A: You can use nth root test for absolute value and obtain absolute convergence.
$$\sqrt[n]{|x|^n(n+1)} = |x|$$
A: $f(x)=\sum_{k=0}^{\infty} x^k=\frac{1}{1+x}$ absolutely converges for $x\in(-1,1)$.  So you can take the derivative term by term to get $f'(x) = \sum_{k=1}^{\infty} kx^{k-1} = \sum_{k=0}^{\infty} (k+1)x^k = -\frac{1}{(1+x)^2}$.  You can use a root test to prove that it converges.  If a power series absolutely converges by the root test, then so will its term by term derivative series.
A: For proving (2), you can do the following:
$$kx^k = x^{k - \log_{1/x}k}$$
Now we only need to show that as $k$ grows, $k - \log_{c} k$ grows without bound as $k$ grows without bound for $c > 1$. For this, consider the function $$g(x) = x - \log_c x$$ from $\mathbb{R}_{>0}$ to itself. We then have
$$g'(x) = 1 - \frac{1}{x \ln c}$$
For $x > x_0 = \frac{1}{2\ln c}$, we have $g'(x) > \frac{1}{2}$. Thus $g(x) > g(x_0) + \frac{1}{2} (x - x_0)$, which shows that $g(x)$ grows without bound as $x$ grows without bound, which was what we needed to show.
A: We claim the limit is $\frac{1}{(1+x)^2}$. It suffices to prove by induction that$$\sum_{k=0}^n(k+1)(-x)^k-\frac{1}{(1+x)^2}=(-x)^n\frac{x(n+2+(n+1)x)}{(1+x)^2},$$which has $n\to\infty$ limit $0$ for $x\in(0,\,1)$. The base step $n=0$ works, as does the inductive step because$$(n+2)(-x)^{n+1}+(-x)^n\frac{x(n+2+(n+1)x)}{(1+x)^2}=(-x)^{n+1}\frac{x(n+3+(n+2)x)}{(1+x)^2}$$(I'll leave you to double-check that).
