For example, with m= 1 that is 1+ 1/2= 3/2. With m= 2 that is 1+ 1+ 1/3= 7/3. With m= 3 that is 1+ 3/2+ 1+ 1/4= 15/4.
But what I would do is note that $\displaystyle (x+ 1)^m= \sum_{n= 0}^m\begin{pmatrix}m \\ n\end{pmatrix}x^n$ and integrating both sides $\displaystyle \frac{1}{m+ 1}(x+ 1)^{m+1}= \sum_{n= 0}^m \frac{1}{n+1}\begin{pmatrix} m\\ n\end{pmatrix}x^{n+1}$. Now let x= 1.
$$\begin{align*}\frac1{n+1}{m \choose n} &= \frac1{n+1} \cdot \frac{m!}{n!(m-n)!} = \frac{m!}{(n+1)!(m-n)!} \\ &= \frac1{m+1}\cdot\frac{(m+1)!}{(n+1)!\big((m+1)-(n+1)\big)!} \\ &= \frac1{m+1}{m+1 \choose n+1}\end{align*}$$
So $$\begin{align*}\sum_{n=0}^{m} \frac1{n+1} {m \choose n} &= \sum_{n=0}^{m} \frac1{m+1}{m+1 \choose n+1} = \frac1{m+1} \sum_{n=1}^{m+1} {m+1 \choose n} \\ &= \frac1{m+1}\left[ \left( \sum_{n=0}^{m+1} {m+1 \choose n} \right) - {m+1 \choose 0} \right] \\ &= \frac1{m+1}\left[ \left( \sum_{n=0}^{m+1} {m+1 \choose n} \right) - 1 \right]\end{align*}$$
Now the summation is simply the sum of all elements in the $(m+1)^\text{th}$ row of Pascal's Triangle (the one with the second element equal to $(m+1)$) and $$ \sum_{n=0}^{m+1} {m+1 \choose n} = (1+1)^{m+1} = 2^{m+1}$$
Thus $$\sum_{n=0}^{m} \frac1{n+1} {m \choose n} = \frac1{m+1}\left[ 2^{m+1} - 1 \right]$$