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$$\sum _{i=1}^n\left(-1\right)^{i+1}\:\binom{n}{i}\:\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$ Can someone give me a hint on how to give the proof, I am stuck when I am proving it for p(n+1):

$$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\left(-1\right)^{n+2}\cdot \frac{1}{n+1}$$

is there a handy way to get rid of the $-1$?

$\sum \:_{i=1}^{n+1}\left(-1\right)^{i+1}\:\binom{n+1}{i}\:\frac{1}{i}=\sum \:_{i=1}^{n+1}\left(-1\right)^{i+1}\:\left[\binom{n}{i}+\binom{n}{i-1}\right] \frac{1}{i}=\sum \:\:_{i=1}^{n+1}\left(-1\right)^{i+1}\binom{n}{i}\frac{1}{i}+\binom{n+1}{i} \frac{1}{n}$

Martin Sleziak
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  • What brings you to this expression with $(-1)^{n+2}$? – Element118 Oct 21 '15 at 11:39
  • change n with n+1 and then take the sum of the first n and change it with the right side of the expression the only one that's left is n+1 term which turns out to be (-1)^n+2... –  Oct 21 '15 at 11:46

1 Answers1

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HINT :

Use $$\binom{n}{i}=\binom{n-1}{i}+\binom{n-1}{i-1}$$ and $$\binom{n-1}{i-1}\frac 1i=\frac 1n\binom{n}{i}.$$

mathlove
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  • $\sum :{i=1}^{n+1}\left(-1\right)^{i+1}:\binom{n+1}{i}:\frac{1}{i}=\sum :{i=1}^{n+1}\left(-1\right)^{i+1}:\left[\binom{n}{i}+\binom{n}{i-1}\right] \frac{1}{i}=\sum ::_{i=1}^{n+1}\left(-1\right)^{i+1}\binom{n}{i}\frac{1}{i}+\binom{n+1}{i} \frac{1}{n}$ –  Oct 21 '15 at 13:25
  • I am lost at this step. –  Oct 21 '15 at 13:32
  • @AleksXPO: $\sum_{i=1}^{n+1}(-1)^{i+1}[\binom{n}{i}+\binom{n}{i-1}]\frac 1i=\sum_{i=1}^{n+1}(-1)^{i+1}\binom{n}{i}\frac 1i+\sum_{i=1}^{n+1}(-1)^{i+1}\binom{n}{i-1}\frac 1i$ $=\sum_{i=1}^{\color{red}{n}}(-1)^{i+1}\binom{n}{i}\frac 1i+\sum_{i=1}^{n+1}(-1)^{i+1}\binom{n+1}{i}\frac{1}{\color{red}{n+1}}$ Can you take it from here? – mathlove Oct 21 '15 at 14:41
  • the first part is equal to 1+1/2+...+1/n , now the problem is how to show that the other part is just 1/n+1.. that's where I was stuck at :/ –  Oct 21 '15 at 17:23
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    @AleksXPO: For the other part, note that by the binomial theorem, we have $\sum_{i=\color{red}{0}}^{n+1}(-1)^{i+1}\binom{n+1}{i}=(-1+1)^{n+1}=0$. So, the other part equals $\frac{1}{n+1}$. – mathlove Oct 21 '15 at 17:29
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    Thanks I get it, I add the i=0 and subtract it which gives +1/n+1 –  Oct 21 '15 at 17:34