Using summation by parts to evaluate an alternating sum

Question

I want to evaluate $$ \sum_{k=0}^n (-1)^k \binom{n}{k} k. $$ I tried summation by parts, i.e. the formula $$ \sum_{k=0}^n (f(k+1) - f(k))g(k) = f(n+1)g(n+1) - f(0)g(0) - \sum_{k=0}^n f(k+1) (g(k+1) - g(k)) $$ with $f(k+1) - f(k) = (-1)^k \binom{n}{k}$ and $g(k) = k$. As $$ (-1)^{k+1}\binom{n-1}{k+1} - (-1)^k \binom{n-1}{k} = (-1)^{k+1} \left( \binom{n-1}{k+1} + \binom{n-1}{k} \right) = (-1)^{k+1} \binom{n}{k+1} $$ we have $f(k) = (-1)^k \binom{n-1}{k}$. Plugging into the formula \begin{align*} \sum_{k=0}^n (-1)^k \binom{n}{k} k & = (-1)^{n+1} \binom{n-1}{n+1} (n+1) - (-1)^0 \binom{n-1}{0}\cdot 0 - \sum_{k=0}^n (-1)^{k+1} \binom{n-1}{k+1} \\ & = \sum_{k=0}^n (-1)^{k} \binom{n-1}{k+1}. \end{align*} But for example if $n = 3$ then $$ \sum_{k=0}^n (-1)^k \binom{n}{k} k = -3 + 6 -3 = 0 $$ but $$ \sum_{k=0}^n (-1)^{k} \binom{n-1}{k+1} = \binom{2}{1} - \binom{2}{2} = 1 $$ which is not equal, but I cannot see whats wrong with the above derivation??

Answer 1

Summation by parts is definitely an overkill, differentiation a lesser overkill. For any $n\geq 1$ we have $$ k\binom{n}{k} = n\binom{n-1}{k-1} $$ and $$ \sum_{k=0}^{n}(-1)^k\binom{n}{k}k = \sum_{k=1}^{n}(-1)^k\binom{n}{k}k = -n\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}=-n\sum_{j=0}^{n-1}\binom{n-1}{j}(-1)^j $$ so your sum is constantly zero for any $n\geq 2$ and you just have to compute it by hand for $n=0$ and $n=1$.

Answer 2

You need $f(k+1) - f(k) = (-1)^k \binom{n}{k}$. But $$ (-1)^{k+1}\binom{n-1}{k+1} - (-1)^k \binom{n-1}{k} = (-1)^{k+1} \left( \binom{n-1}{k+1} + \binom{n-1}{k} \right) = (-1)^{k+1} \binom{n}{k+1} $$

So

$$ (-1)^{k}\binom{n-1}{k} - (-1)^{k-1} \binom{n-1}{k-1} = (-1)^{k} \left( \binom{n-1}{k} + \binom{n-1}{k-1} \right) = (-1)^{k} \binom{n}{k} $$

$f(k)=(-1)^{k-1} \binom{n-1}{k-1}$.

Of course, When $k=0$ and $k=n+1$, we have special cases and should consider seperately.

$f(1)=\binom{n-1}{0}=1$

$f(1)-f(0)=(-1)^0\binom{n}{0}=1$ and so $f(0)=0$

$f(n)=(-1)^{n-1}\binom{n-1}{n-1}=(-1)^{n-1}$

$f(n+1)-f(n)=(-1)^n\binom{n}{n}=(-1)^n$ and so $f(n+1)=(-1)^n+(-1)^{n-1}=0$

\begin{align*} \sum_{k=0}^n (-1)^k \binom{n}{k} k & =(0) (n+1) -(0)(0) - \sum_{k=1}^{n-1} (-1)^{k} \binom{n-1}{k}-f(1)(g(1)-g(0))-f(n+1)(g(n+1)-g(n)) \\ & = - \sum_{k=0}^{n-1} (-1)^{k} \binom{n-1}{k}+(-1)^0\binom{n-1}{0}-1\\ &=0 \end{align*}

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My work:

$\displaystyle (1+x)^n=\sum_{k=1}^n\binom{n}{k}x^k+1$

Differentiating,

$\displaystyle n(1+x)^{n-1}=\sum_{k=1}^nk\binom{n}{k}x^{k-1}$

Put $x=-1$.

\begin{align*} \sum_{k=1}^nk\binom{n}{k}(-1)^{k-1}&=0\\ \sum_{k=1}^nk\binom{n}{k}(-1)^{k}&=(-1)(0)\\ \sum_{k=0}^nk\binom{n}{k}(-1)^{k}&=0+0=0\\ \end{align*}