summation for $x \gt 1$

Question

how do I show that the $\sum_{n=0}^{n=N} (n+1)x^n$ is less than or equal to the square of this whole sum : $\sum_{n=0}^{n=N}x^n$

Tried induction didn't work. Got messy

tried doing directly by using the formula to a geometric series- didn't work

Would really appreciate some help

Thanks

Answer 1

Hint: Both sums can be done in closed form. One is a geometric series, the other is the derivative of a geometric series.

Answer 2

You can just get the close form sums of this two sequences. The difference is something positive.

Answer 3

$$\begin{array}\\ \displaystyle\left(\sum_{n=0}^N x^n\right)^2 &=\displaystyle\sum_{n=0}^N \sum_{m=0}^N x^{n+m}\\ &\ge\displaystyle\sum_{n=0}^N \sum_{m=0}^{N-n} x^{n+m}\\ &=\displaystyle\sum_{j=0}^N x^j \sum_{m=0}^{j} 1 \quad (j=n+m)\\ &=\displaystyle\sum_{j=0}^N (j+1)x^j\\ \end{array} $$