- #1
gruba
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Homework Statement
Prove that [itex]\forall n\in \mathbb{N}[/itex]
[tex]\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}[/tex]
Homework Equations
-Induction
-Summation
-Binomial coefficient
The Attempt at a Solution
For [itex]n=1[/itex] equality is true.
For [itex]n=m[/itex]
[itex]m-{m\choose 2}\frac{1}{2}+...+(-1)^{m+1}\frac{1}{m}=1+\frac{1}{2}+...+\frac{1}{m}[/itex]
For [itex]n=m+1[/itex]
[itex]\left(\sum\limits_{k=1}^{m}(-1)^{k+1}{m\choose k}\frac{1}{k}\right)+(-1)^{m+2}\frac{1}{m+1}=1+\frac{1}{2}+...+\frac{1}{m+1}[/itex]
If [itex]m[/itex] is even, equality is true, but not if [itex]m[/itex] is odd.
Is this correct?
Could someone explain this in detail?