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Prove an identity of given finite sums

Prove the identity:

    \[ \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} = \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{k+n+1}. \]

Proof. Using the hint (that \frac{1}{k+m+1} = \int_0^1 t^{k+m} \, dt) we start with the expression on the left,

    \begin{align*}  \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} &= \sum_{k=0}^n (-1)^k \binom{n}{k} \int_0^1 t^{m+k} \, dt \\[10pt]  &= \sum_{k=0}^n \int_0^1 (-1)^k \binom{n}{k} t^m t^k \, dt \\[10pt]  &= \int_0^1 \sum_{k=0}^n (-1)^k \binom{n}{k} t^m t^k \, dt &(\text{finite sum}) \\[10pt]  &= \int_0^1 t^m \sum_{k=0}^n \binom{n}{k} (-t)^k \, dt \\[10pt]  &= \int_0^1 t^m \sum_{k=0}^n \binom{n}{k} (-t)^k (1)^{n-k} \, dt \\[10pt]  &= \int_0^1 t^m (1-t)^n \, dt &(\text{Binomial theorem}). \end{align*}

(The interchange of the sum and integral is fine since it is a finite sum. Those planning to take analysis should note that this cannot always be done in the case of infinite sums.) Now, we have a reasonable integral, but we still want to get everything back into the form of the sum on the right so we make the substitution s = 1-t, ds = -dt. This gives us new limits of integration from 1 to 0. Therefore, we have

    \begin{align*}  \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} &= \int_0^1 t^m (1-t)^n \, dt \\[10pt]  &= -\int_1^0 (1-s)^m s^n \, ds \\[10pt]  &= \int_0^1 (1-s)^m s^n \, ds \\[10pt]  &= \int_0^1 s^n \sum_{k=0}^m \binom{m}{k} (-s)^k 1^{m-k} \, ds &(\text{Binomial theorem})\\[10pt]  &= \int_0^1 \sum_{k=0}^m \binom{m}{k} (-1)^k s^{k+n} \, ds \\[10pt]  &= \sum_{k=0}^m (-1)^k \binom{m}{k} \int_0^1 s^{k+n} \, ds \\[10pt]  &= \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{k+n+1}. \qquad \blacksquare \end{align*}

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