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My textbook represents an alternating series as an infinite series of the form

$$ \sum_{n=1}^{\infty}(-1)^{n-1}a_n = a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + ...$$ where $ a_1, a_2 ... $ is a sequence with $a_n \ge 0$ for all $n$.

However, I was wondering if it would be correct to also represent an alternative series in the form

$$ \sum_{n=1}^{\infty}(-1)^{n+1}a_n = a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + ...$$ where $ a_1, a_2 ... $ is a sequence with $a_n \ge 0$ for all $n$?

Notice the $n+1$.

It seems to me that this would be identical.

Thank you.

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    $\begingroup$ Yes it will be same ; $\endgroup$ – Learnmore Nov 10 '16 at 3:16
  • $\begingroup$ @learnmore Thank you. I would appreciate if someone could verify that this response is correct. :) $\endgroup$ – The Pointer Nov 10 '16 at 3:33
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    $\begingroup$ Hint: $(-1)^{n+1} \,/\, (-1)^{n-1} = (-1)^{(n+1)-(n-1)}=(-1)^2 =1$ so the two are always equal. $\endgroup$ – dxiv Nov 10 '16 at 3:38
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Since the -1 is only used to determine the sign of $$ a_1,a_2,...$$ the exponent's value doesn't matter as long as it's either odd or even respectively.

In other words,

doing $(-1)^{1+1}$ or $(-1)^{1-1}$ both gives a positive value to -1

while $(-1)^{2+1}$ or $(-1)^{2-1}$ both gives a negative value to -1

Extra question you can think about : What about those, are they the same? $(-1)^{2n+1}$ $(-1)^{2n}$

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