find the sum of the following where $n \ge 1 $
$\sum_{k=1}^{n} (-1)^{k+1} \frac{1}{k} \binom{n}{k}$
i know that $\binom {n}{k} = \frac{n!}{(n-k)! k!}$
Now $\sum_{k=1}^{n} (-1)^{k+1} \frac{1}{k} \binom{n}{k}=\sum_{n=0}^{k} (-1)^{k+1}\frac{1}{k} \frac{n!}{(n-k)! k!}$
By leibnitz test(alternating series) that is $\sum_{k=1}^{n} (-1)^{k+1} \frac{1}{k} \binom{n}{k}= 0$
Is im correct or not ??
Pliz verified me....
thanks u
$$\sum_{k=1}^n(-1)^{k+1}\frac1k\binom nk =\sum_{k=1}^n(-1)^{k+1}\binom{n}{k} \int_0^1 x^{k-1}\,dx=\int_0^1\frac{1-(1-x)^n}{x}\,dx =\int_0^1\frac{1-y^n}{1-y}\,dy.$$ You can reduce this last integrand to a polynomial and integrate termwise.
Hint:
$$(1-x)^n=\sum_{k=0}^n\binom{n}{k}(-1)^kx^k\tag{1}$$
Can you manipulate this sum using differentiation/integration?
I take it OP is satisfied, so I'll expand on my answer:
To get OP's sum we will divide $(1)$ by $x$ and integrate it, however not every term in $(1)$ is divisible by $x$. Instead it is easier to consider: $$(1-x)^n-1=\sum_{k=1}^n\binom{n}{k}(-1)^kx^k$$ Then the rest will fall out:
$$\int_0^1\frac{(1-t)^n-1}{t}dt=\int_0^1\sum_{k=1}^n\binom{n}{k}(-1)^kt^{k-1}dt=\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k}$$ The integral on the left can be evaluated by a simple substitution $u=1-t$, this gives $$-\int_{1}^{0}\frac{u^n-1}{1-u}du=\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k}\\\int_0^1\frac{1-u^n}{1-u}du=-\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k}=\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k+1}}{k}$$ We also know that $$\frac{1-u^n}{1-u}=1+u+u^2+\dots+u^{n-1}$$ We find that $$\int_0^1 1+u+u^2+\dots+u^{n-1}du=\sum_{k=1}^{n}\frac{1}{k}=H_n$$
