summation algebra for $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1}$

Question

Why does $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1} = 1 + 2\sum_{n=1}^\infty x^n$? Shouldn't this be $1 + x + 2\sum_{n=1}^\infty x^n$ because of the $n+1$ in the second summation?

Answer 1

$$\sum_{n=0}^{\infty}x^{n+1}$$ setting $k=n+1$ we have the following: when $n=0$, $k=1$ and when $n \rightarrow \infty$ , $k \rightarrow \infty$ $$\sum_{n=0}^{\infty}x^{n+1}=\sum_{k=1}^{\infty}x^{k}$$ Then you can change again the symbol to get $$\sum_{n=0}^{\infty}x^{n+1}=\sum_{n=1}^{\infty}x^{n}$$ For the other sum we have the following:

$$\sum_{n=0}^{\infty}x^n=x^0+\sum_{n=1}^{\infty}x^n=1+\sum_{n=0}^{\infty}x^n$$

Answer 2

$$\sum_{n=0}^\infty x^n=1+\sum_{n=1}^\infty x^n \ \ and\ \ \sum_{n=0}^\infty x^{n+1}=\sum_{n=1}^\infty x^{n} $$

Answer 3

Hint:

If you're uncertain how shifting indices affects a series, it can be helpful to introduce a different variable and work through it like a substitution problem. For instance, substituting $n=m-1$, we see:

$$\sum_{n=0}^{\infty}x^{n+1}=\sum_{m-1=0}^{\infty}x^{m-1+1}=\sum_{m=1}^{\infty}x^{m}=\sum_{n=1}^{\infty}x^{n}.$$

And at the end we simply relabel the dummny index $m$ back to $n$.

Answer 4

This may convince you.

\begin{array}\ \displaystyle\sum_{n=0}^{\infty} x^n \hphantom{+ \displaystyle\sum_{n=0}^{\infty} x^{n+1}} &= 1 + \hphantom{2}x + \hphantom{2}x^2 + \dots\\ &\\ \hphantom{\displaystyle\sum_{n=0}^{\infty} x^n +} \displaystyle\sum_{n=0}^{\infty} x^{n+1} &= \hphantom{1 + 2}x + \hphantom{2}x^2 + \dots\\ &\\ \displaystyle\sum_{n=0}^{\infty} x^n + \sum_{n=0}^{\infty} x^{n+1} &= 1 + 2x + 2x^2 + \dots \end{array}