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Prove that:

$$\sum_{j=1}^{n}(-1)^j\binom{n}{j}\frac{1}{j+1}=-\frac{n}{n+1} $$

I know that

1.I need prove it by induction
2.this can be helpful :

$ n\binom{n-1}{j-1}=j\binom{n}{j} $

whatever, i tried to do it by induction and i don´t know how i can to go from $\sum_{j=1}^{n-1}(-1)^j\binom{n-1}{j}\frac{1}{j+1} $ to $\sum_{j=1}^{n}(-1)^j\binom{n}{j}\frac{1}{j+1} $

I mind, I know that

$\sum_{j=1}^{n}(-1)^j\binom{n}{j}\frac{1}{j+1} = \sum_{j=1}^{n-1}(-1)^j\binom{n-1}{j}\frac{n}{n-j}\frac{1}{j+1} + (-1)^n\frac{1}{n+1} $ But I can´t continue...

thank for your help

Salech Alhasov
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Luis GC
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2 Answers2

3

Hint: You may want to apply $${n+1\choose j+1}=\frac{n+1}{j+1}{n\choose j}$$

Shuchang
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Multiply both sides by $n + 1$. $$\sum_{j=1}^{n}(-1)^j\binom{n+1}{j+1}=-n$$ And now:

$\sum_{j=1}^{n}(-1)^j\binom{n+1}{j+1}=\sum_{j=0}^{n}(-1)^j\binom{n+1}{j+1}-(n+1)=\left(\sum_{j=0}^{n+1}(-1)^{j+1}\binom{n+1}{j}+1\right)-(n+1)=0+1-n-1=-n$

This term $\sum_{j=0}^{n+1}(-1)^{j+1}\binom{n+1}{j}$ is an alternating sum and difference of binomial coefficients and is equal to zero.

However, I didn't use mathematical induction in this proof.

eudoxyz
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