Summation of alternating series:$\sum_{k=1}^{n} (-1)^{k-1}k$

Question

Alternate summations $S_1=1-2+3-4+5-......+(2m-1)$ and $S_2=1-2+3-4+5-......-2m$ can be found as $\pm m$, respectively by arranging $$S_1=[1+2+3+4+5+.....+(2m-1)]-4[1+2+3+4+....+m]$$ We can get the summation of alternating series: $$F=\sum_{k=1}^{n} (-1)^{k-1} k~~~~~~~~(*)$$ as $$G=\sum_{k=1}^{n}k x ^{k-1}=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x} \implies F=\frac{1-(-1)^n(2n+1)}{4}.$$ The question is: How to get $F(*)$ by some other method like Telescopic summation etc.?

Answer 1

The sum sequence is

$S(n) = \displaystyle \sum_{k = 0}^n (-1)^{k-1} k $

which can be thought of as

$S(n) = \displaystyle \sum_{k=0}^n k r^{k-1} $

Evaluated at $r = -1$. On the other hand, we can write this sum as

$S(n) = \displaystyle \dfrac{d}{dr} \left( \sum_{k=0}^n r^{k} \right) $

And this simplifies to

$S(n) = \displaystyle \dfrac{d}{dr} \left( \dfrac{ 1 - r^{n+1} }{1 - r} \right)$

Taking the derivative,

$S(n) = \dfrac{ -(n+1) r^n (1 - r) + (1 - r^{n+1} ) }{ (1 - r)^2} $

which simplifes to

$S(n) = \dfrac{ 1 + r^n ( - n - 1 + n r ) }{ (1 - r)^2} $

Using $r = -1$ , this becomes

$S(n) = \dfrac{ 1 + (-1)^{n+1} ( 2n + 1 ) }{ 4 } $

To verify the formula, take $n = 5 $, then $S(5) = 1 - 2 + 3 - 4 + 5 = 3 $

And from the formula, $ S(5) = \dfrac{ 1 + (-1)^6 (11) } {4} = 3 $

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Another method one can apply is taking the terms of the sequence S(n), then

$(E - 1) S(n) = a(n) $

where $E$ is the advance operator, i.e. $E(S(n)) = S(n+1) $, and $ a(n) = (-1)^{(n-1)} n $

The annihilator of $a(n)$ is $ (E + 1)^2 $, therefore, we now have

$ (E - 1) (E + 1)^2 S(n) = 0 $

The roots of the characteristic equation are clearly, $1, -1, -1$. Therefore,

$S(n) = A + B (-1)^n + C n (-1)^n $

To determine $A, B, C$ , use the first three terms of $S(n)$

$S(1) = 1 = A - B - C $

$S(2) = -1 = A + B + 2 C $

$S(3) = 2 = A - B - 3 C $

Solving this $3 \times 3$ linear system of equations, gives us

$A = \dfrac{1}{4} , B = - \dfrac{1}{4}, C = -\dfrac{1}{2} $

Hence,

$S(n) = \dfrac{ 1 - (-1)^n - 2 n (-1)^n }{4} = \dfrac{ 1 - (-1)^n (1 + 2n) }{4} $

And this is exactly what we got earlier.