sequences and series - I simply want verification that $sum_{2}^{infty} left(frac{1}{2}right)^{x-1} = sum_{1}^{infty} left(frac{1}{2}right)^{x}$?

Sorry to bother. Up late studying and want to know for sure that $\sum_{2}^{\infty} \left(\frac{1}{2}\right)^{x-1} = \sum_{1}^{\infty} \left(\frac{1}{2}\right)^{x}$?

It sure seems that way since:

$$\sum_{2}^{\infty} \left(\frac{1}{2}\right)^{x-1} = \left(\frac{1}{2}\right)^{2-1} + \left(\frac{1}{2}\right)^{3-1} + \left(\frac{1}{2}\right)^{4-1} + \left(\frac{1}{2}\right)^{5-1} + \cdots = \left(\frac{1}{2}\right)^{1} + \left(\frac{1}{2}\right)^{2} + \left(\frac{1}{2}\right)^{3} + \left(\frac{1}{2}\right)^{4} + \cdots$$

Which would be equal to
$\sum_{1}^{\infty} \left(\frac{1}{2}\right)^{x}$

Thanks!

Idle

Answer

In the current version of the question it is correct because it is just an index shift. Both sides of the equality converge absolutely and you can identify the terms on one side with the terms on the other. It is just a shift of the index by $1$. The sum is $1$.