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Answer not correct. It has to be revised....

Since we have $r\geq k-1$ it is convenient to set $r=k-1+l$ with $l\geq 0$.

We obtain \begin{align*} &\color{blue}{I(k,k-1+l)}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=1}^{l+1}(-1)^{l-j}\binom{k-1+l-j}{k-2}\frac{1}{x^j}+\frac{1}{(1+x)^{k-1}x^{l+1}}\right\}\tag{1}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{l-j+1}\binom{k+l-j-2}{l-j}\frac{1}{x^{j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{-(k-1)}{j}x^j\right\}\tag{2}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{j+1}\binom{k+j-2}{j}\frac{1}{x^{l-j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{k+j-2}{j}(-x)^j\right\}\tag{3}\\ &\quad=\lim_{x\to 0}\left\{\frac{1}{x^{l+1}}\sum_{j=l+1}^\infty\binom{k+j-2}{j}(-x^j)\right\}\tag{4}\\ &\quad\,\,\color{blue}{=(-1)^{l+1}\binom{k+l-1}{l+1}}\tag{5} \end{align*}

Comment:

• In (1) we set $r=k-1+l$ in $I(k,r)$.

• In (2) we shift the index of the sum by one to start with $j=0$ and we use the binomial series expansion.

• In (3) we reverse the order of summation in the finite sum by setting $j\to l-j$. We also apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we see the $l+1$ terms of the sum cancel. The series now starts with $j=l+1$.

• In (5) we apply the limit and all terms cancel besides the term with $x^0$.

We finally conclude from (5) by using $r=k-1+l$ \begin{align*} \color{blue}{I(k,r)}&=(-1)^{r-k}\binom{r}{r-k+2}\color{blue}{=(-1)^{r-k}\binom{r}{k-2}} \end{align*}

Note: The result is in accordance with $OP's$ example: $$I(k,k)=\binom{k}{k-2}=\binom{k}{2}=\frac{k(k-1)}{2}$$

Since we have $r\geq k-1$ it is convenient to set $r=k-1+l$ with $l\geq 0$.

We obtain \begin{align*} &\color{blue}{I(k,k-1+l)}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=1}^{l+1}(-1)^{l-j}\binom{k-1+l-j}{k-2}\frac{1}{x^j}+\frac{1}{(1+x)^{k-1}x^{l+1}}\right\}\tag{1}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{l-j+1}\binom{k+l-j-2}{l-j}\frac{1}{x^{j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{-(k-1)}{j}x^j\right\}\tag{2}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{j+1}\binom{k+j-2}{j}\frac{1}{x^{l-j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{k+j-2}{j}x^j\right\}\tag{3}\\ &\quad=\lim_{x\to 0}\left\{\frac{1}{x^{l+1}}\sum_{j=l+1}^\infty\binom{k+j-2}{j}x^j\right\}\tag{4}\\ &\quad\,\,\color{blue}{=\binom{k+l-1}{l+1}}\tag{5} \end{align*}

Comment:

• In (1) we set $r=k-1+l$ in $I(k,r)$.

• In (2) we shift the index of the sum by one to start with $j=0$ and we use the binomial series expansion.

• In (3) we reverse the order of summation in the finite sum by setting $j\to l-j$. We also apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we see the $l+1$ terms of the sum cancel. The series now starts with $j=l+1$.

• In (5) we apply the limit and all terms cancel besides the term with $x^0$.

We finally conclude from (5) by using $r=k-1+l$ \begin{align*} \color{blue}{I(k,r)}&=\binom{k+(r-k+1)-1}{r-k+2}=\binom{r}{r-k+2}\color{blue}{=\binom{r}{k-2}} \end{align*}

Note: The result is in accordance with $OP's$ example: $$I(k,k)=\binom{k}{k-2}=\binom{k}{2}=\frac{k(k-1)}{2}$$

Answer not correct. It has to be revised....

Since we have $r\geq k-1$ it is convenient to set $r=k-1+l$ with $l\geq 0$.

We obtain \begin{align*} &\color{blue}{I(k,k-1+l)}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=1}^{l+1}(-1)^{l-j}\binom{k-1+l-j}{k-2}\frac{1}{x^j}+\frac{1}{(1+x)^{k-1}x^{l+1}}\right\}\tag{1}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{l-j+1}\binom{k+l-j-2}{l-j}\frac{1}{x^{j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{-(k-1)}{j}x^j\right\}\tag{2}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{j+1}\binom{k+j-2}{j}\frac{1}{x^{l-j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{k+j-2}{j}x^j\right\}\tag{3}\\ &\quad=\lim_{x\to 0}\left\{\frac{1}{x^{l+1}}\sum_{j=l+1}^\infty\binom{k+j-2}{j}x^j\right\}\tag{4}\\ &\quad\,\,\color{blue}{=\binom{k+l-1}{l+1}}\tag{5} \end{align*}

Comment:

• In (1) we set $r=k-1+l$ in $I(k,r)$.

• In (2) we shift the index of the sum by one to start with $j=0$ and we use the binomial series expansion.

• In (3) we reverse the order of summation in the finite sum by setting $j\to l-j$. We also apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we see the $l+1$ terms of the sum cancel. The series now starts with $j=l+1$.

• In (5) we apply the limit and all terms cancel besides the term with $x^0$.

We finally conclude from (5) by using $r=k-1+l$ \begin{align*} \color{blue}{I(k,r)}&=\binom{k+(r-k+1)+1}{r-k+2}=\binom{r}{r-k+2}\color{blue}{=\binom{r}{k-2}} \end{align*}

Note: The result is in accordance with $OP's$ example: $$I(k,k)=\binom{k}{k-2}=\binom{k}{2}=\frac{k(k-1)}{2}$$

Since we have $r\geq k-1$ it is convenient to set $r=k-1+l$ with $l\geq 0$.

We obtain \begin{align*} &\color{blue}{I(k,k-1+l)}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=1}^{l+1}(-1)^{l-j}\binom{k-1+l-j}{k-2}\frac{1}{x^j}+\frac{1}{(1+x)^{k-1}x^{l+1}}\right\}\tag{1}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{l-j+1}\binom{k+l-j-2}{l-j}\frac{1}{x^{j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{-(k-1)}{j}x^j\right\}\tag{2}\\ &\quad=\lim_{x\to 0}\left\{\sum_{j=0}^l(-1)^{j+1}\binom{k+j-2}{j}\frac{1}{x^{l-j+1}} +\frac{1}{x^{l+1}}\sum_{j=0}^\infty\binom{k+j-2}{j}x^j\right\}\tag{3}\\ &\quad=\lim_{x\to 0}\left\{\frac{1}{x^{l+1}}\sum_{j=l+1}^\infty\binom{k+j-2}{j}x^j\right\}\tag{4}\\ &\quad\,\,\color{blue}{=\binom{k+l-1}{l+1}}\tag{5} \end{align*}

Comment:

• In (1) we set $r=k-1+l$ in $I(k,r)$.

• In (2) we shift the index of the sum by one to start with $j=0$ and we use the binomial series expansion.

• In (3) we reverse the order of summation in the finite sum by setting $j\to l-j$. We also apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we see the $l+1$ terms of the sum cancel. The series now starts with $j=l+1$.

• In (5) we apply the limit and all terms cancel besides the term with $x^0$.

We finally conclude from (5) by using $r=k-1+l$ \begin{align*} \color{blue}{I(k,r)}&=\binom{k+(r-k+1)-1}{r-k+2}=\binom{r}{r-k+2}\color{blue}{=\binom{r}{k-2}} \end{align*}

Note: The result is in accordance with $OP's$ example: $$I(k,k)=\binom{k}{k-2}=\binom{k}{2}=\frac{k(k-1)}{2}$$