Question

How to change the exponent of z to -n?
\(\sum_{n=0}^\infty 2^{n-1}((-1)^n +1)z^{-(n+1)}\)
Please, help

Answer 1

it is right? \(\sum_{n=1}^\infty 2^{n-2}((-1)^{n-1}+1)z^{-n}\)

Answer 2

\[
\sum_{n=1}^\infty 2^{n-1}((-1^n+1)z^{-(n+1)}=z^{-1}\sum_{n=1}^\infty 2^{n-1}((-1^n+1)z^{-n}
\]You look like you just had a massive brain fart.

Answer 3

Actually me too.
\[
\sum_{n=1}^\infty 2^{n-1}((-1)^n+1)z^{-(n+1)}=z^{-1}\sum_{n=1}^\infty 2^{n-1}((-1)^n+1)z^{-n}
\]

Answer 4

in the first sum, the lower limit is n =0

Answer 5

Hence, if we use k is a new letter in second limit, then we change n+1 =k
so that \(z ^{-(n+1)}= z^{-k}\)
for lower limit, if n+1 =k and n =0, then k =1, the limit of the sum is \(\sum_{k=1}^\infty\)
For \(2^{n-1}\), n +1=k, then n = k -1, then n-1 = k-1-1=k-2. So that \(2^{n-1}=2^{k-2}\)
for (-1) ^n, just replace n = k-1, hence it becomes \((-1)^{k-1}\)
combine all: \(\sum_{k=1}^\infty 2^{k-1}((-1)^{k-1}+1)z^-k\)

Answer 6

sorry for the last one of 2^. It is \(2^{k-2}\)

Answer 7

You sneaky index-changing bastard!!!! jk
\[
\sum_{n=0}^\infty 2^{n-1}((-1)^n+1)z^{-(n+1)}=(2z)^{-1}\sum_{n=0}^\infty 2^n((-1)^n+1)z^{-n}
\]I would prefer not to do any index shift. This some is also equal to \[z^{-1}\sum_{n=0}^\infty 2^{2n}z^{-2n}\] since \((-1)^n+1=0\) for odd n and 2 for even n.