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I have $\sum_{n=0}^\infty \frac{{(-1)}^n}{z^{n+1}}$ now I want to change the lower limit to begin at $1$ instead of $0$. So I let $k=n+1$, and I'm supposed to get $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{z^k}$. I know how to get it for $z$ because I plugged in $n=k-1$, but when I plug it in for $(-1)^n$ I get $(-1)^{k-1}$. I know it's the same at the end, but did I make a mistake?

Thanks

A A
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1 Answers1

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Not to mess up with powers of $-1$, let us suppose that you have $\sum_{n=0}^\infty a_nb_{n+1}$, then changing the summation variable $k = n+1$ leads to $$ \sum_{k=1}^\infty a_{k-1}b_k $$ as much simple, as it is with a change of vairables in integrals which leads to the corresponding change of limits of integration.

SBF
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  • Thanks, So I changed it correctly right? – A A Mar 28 '13 at 15:47
  • @AA: strictly speaking, you have $$ \sum_{n=0}^\infty (-1)^n \frac{1}{z^{n+1}} = \sum_{k=1}(-1)^{k{\color{red}-}1}\frac{1}{z^k} $$ – SBF Mar 28 '13 at 15:49
  • Yeah, that's what I originally had, but in the solution it was $k+1$ – A A Mar 28 '13 at 15:51
  • @AA: as you noticed, in case $a_n = (-1)^n$ both formulas are equivalent, also having $+1$ in the index is always "safer" since you more rarely leave the domain of the definition (or perhaps, in this case it's author's personal preference). If you agree that the formula I provided for the change of summation variable is correct (is starts with $a_0b_1$ rather than $a_2b_1$ in case of $+1$), then we shall agree that the statement in the book is correct but a bit confusing. – SBF Mar 28 '13 at 15:56
  • Agreed, thank you for the clarification :) – A A Mar 28 '13 at 15:58