Hello everyone how can I calculate the sum of:
$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}$
I tried to use pascal identity and the binomial theorem and didn't success.
Someone can help me please?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Set $n=m-1$ and use math.stackexchange.com/questions/94719/… $\endgroup$ – lab bhattacharjee May 9 '20 at 13:17
-
$\begingroup$ same question $\endgroup$ – Light Yagami Jan 3 at 7:05
Add a comment
|
$\begingroup$
$\endgroup$
Notice that : $${n \choose k}\frac{1}{k+1}={n+1 \choose k+1}\frac{1}{n+1}$$ The sum is $$\sum_{k=0}^{n}{n+1 \choose k+1}\frac{1}{n+1}=\frac{2^{n+1}-1}{n+1}$$
Another possible solution is $$(1+x)^n=\sum_{k=0}^{n}{n \choose k}x^k$$ We can integrate both sides to get $$\int_0^1(1+x)^ndx=\sum_{k=0}^n{n \choose k}\int_0^1x^kdx=\sum_{k=0}^n{n \choose k}\frac{1}{k+1}$$ The LHS is simply $\frac{2^{n+1}-1}{n+1}$ which gives the final result. Hope this helps.
