Prove the identity:
![\[ \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} = \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{k+n+1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0090315e928c0b64b460d908eb30e9f5_l3.png "Rendered by QuickLaTeX.com")
Proof. Using the hint (that 
) 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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-100d1fa891553f7e4dbc15d4c95a8324_l3.png "Rendered by QuickLaTeX.com")
(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 
, 
. 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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0ca8c1fe1cc5bdbf2f7a40415d7baa18_l3.png "Rendered by QuickLaTeX.com")