Proof of an inequality to bound a sequence

Question

I have the sequence $v_n=\sum_{k=1}^{n}\frac{x^k}{k}$, with $\mid x\mid<1$. I'm trying to prove that for $p\geq1$ : $$\mid v_{n+p}-v_n\mid\leq\frac{\mid x\mid^{n+1}}{1-\mid x \mid}$$ What I found, and I'm not sure it's helpful, is that : \begin{align} \mid v_{n+p}-v_n\mid &=\left|\sum\limits_{k=1}^{n+p}\frac{x^k}{k} -\sum\limits_{k=1}^n\frac{x^k}{k} \right|\\ &=\left| \sum\limits_{k=n+1}^{n+p}\frac{x^k}{k} \right|\\ &\leq \sum\limits_{k=n+1}^{n+p}\frac{\mid x\mid^k}{k} \end{align} Each term in the last sum is smaller than $|x|^k$ and there are $p$ terms so the last sum is smaller then $\sum\limits_{k=1}^p|x|^k$ but then again not sure this helps because if you applies the geometric series formula, you still have a $p$ in the answer and not $n+1$. So if anyone could give me a hint on where to start, where to go, but not the full solution I'd appreciate it.

Answer 1

You can follow on like that

$$\begin{align} \mid v_{n+p}-v_n\mid &=\left|\sum\limits_{k=1}^{n+p}\frac{x^k}{k} -\sum\limits_{k=1}^n\frac{x^k}{k} \right|\\ &=\left| \sum\limits_{k=n+1}^{n+p}\frac{x^k}{k} \right|\\ &\leq \sum\limits_{k=n+1}^{n+p}\frac{\mid x\mid^k}{k}\\ &\le \vert x \vert^{n+1}\sum\limits_{k=n+1}^\infty\frac{\mid x\mid^{k-n-1}}{k}\\ &\le \vert x \vert^{n+1}\sum\limits_{k=n+1}^\infty\mid x\mid^{k-n-1}\\ &= \vert x \vert^{n+1} \sum\limits_{k=0}^\infty \vert x \vert^k\\ &= \frac{\vert x \vert^{n+1}}{1-\vert x \vert} \end{align}$$

Answer 2

You have $$ \sum\limits_{k=n+1}^{n+p}\frac{\lvert x\rvert^k}{k} \leq \sum\limits_{k=n+1}^{n+p}\lvert x\rvert^k = \lvert x\rvert^{n+1}\cdot \frac{1-\lvert x\rvert^p}{1-\lvert x\rvert} $$ Now use that $1-\lvert x\rvert^p \leq 1$.