Prove the following Combinatorial identity

Question

For any m $\ge$ 1. Prove that $$\sum_{j=0}^m\frac{1}{j+1}{m \choose j} = \frac{1}{m+1}(2^{m+1} -1 )$$

Attempt: I tried using subsets after rearranging the equation. Then even though the R.H.S gives the no of substes of n+1 excluding empty. I am unable to get a interpretation for L.H.S

Answer 1

HINT:

$$\frac1{j+1}\binom{m}j=\frac1{m+1}\binom{m+1}{j+1}$$

(This is an identity worth knowing; it comes up quite often.)

Answer 2

Start with $$ (1+x)^m=\sum_{j=0}^{m}{m\choose j}x^j $$ and integrate both sides from $0$ to $1$.