B How can I show the sum results in this?

[howabout1337]

$\sum_{n=0}^\infty \frac 1{p^{nz}}=1+\frac1{p^z}+\frac1{p^{2z}}+\frac1{p^{3z}}...$
$\frac 1{p^z}\sum_{n=0}^\infty \frac 1{p^{nz}}=\frac1{p^z}+\frac1{p^{2z}}+\frac1{p^{3z}}+\frac1{p^{4z}}...$
something happens and it shows:
$\frac 1{p^z}\sum_{n=0}^\infty \frac 1{p^{nz}}=\sum_{n=0}^\infty\frac1{p^{nz}}-1$ <== How can I get here from above

 

[Chestermiller]

$\sum_{n=0}^\infty \frac 1{p^{nz}}=1+\frac1{p^z}+\frac1{p^{2z}}+\frac1{p^{3z}}...$
$\frac 1{p^z}\sum_{n=0}^\infty \frac 1{p^{nz}}=\frac1{p^z}+\frac1{p^{2z}}+\frac1{p^{3z}}+\frac1{p^{4z}}...$
something happens and it shows:
$\frac 1{p^z}\sum_{n=0}^\infty \frac 1{p^{nz}}=\frac1{p^{nz}}-1$ <== How can I get here from above

You can't. There should be a summation on the first term on the right hand side.

 

[howabout1337]

You can't. There should be a summation on the first term on the right hand side.

yes, I am sorry. Let me fix that

 

[dRic2]

Subtract 1 from each side of the first identity

 

[howabout1337]

Subtract 1 from each side of the first identity

Perfect. Thank you so much!