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$$\sum^{n}_{i=1} {n \choose i}(-1)^{i+1}\left(\frac{1}{i}\right)=\sum^{n}_{i=1}\frac{1}{i}$$

I want to prove this inductively. For the sake of solving the question, I'm going to ignore formalities. Assume

$$\sum^{k}_{i=1} {k \choose i}(-1)^{i+1}(\frac{1}{i})=\sum^{k}_{i=1}\frac{1}{i}$$

Then

\begin{align} &\sum^{k+1}_{i=1} {k+1 \choose i}(-1)^{i+1}(\frac{1}{i})\\ =&\sum^{k+1}_{i=1} ({k \choose i}+{k \choose i-1})(-1)^{i+1}(\frac{1}{i}) \\ =&\sum^{k+1}_{i=1} {k \choose i}(-1)^{i+1}(\frac{1}{i})+\sum^{k+1}_{i=1} {k \choose i-1})(-1)^{i+1}(\frac{1}{i}) \\ =&\underbrace{\sum^{k}_{i=1} {k \choose i}(-1)^{i+1}(\frac{1}{i})}_{\text{Can the $k+1$ term be ignored since it's undefined?}}+\sum^{k}_{i=0} {k \choose i}(-1)^{i+2}(\frac{1}{i+1}) \\ =&\sum^{k}_{i=1}\frac{1}{i}+ \underbrace{\sum^{k}_{i=0} {k \choose i}(-1)^{i}(\frac{1}{i+1})}_{\text{How do I go about showing that this is $\frac{1}{k+1}$?}} \\ \end{align}

My intuition tells me that the underbraced sum has something to do with a binomial expansion, however that would require the $\frac{1}{1+i}$ term to have an exponent of $k-i$. Any tips on how to go further?

RobPratt
  • 50,938

1 Answers1

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$${k+1\choose i}={k\choose i}+{k\choose i-1}$$ if and only if $1\leq i\leq k$. Therefore you may want to separate the $(k+1)$th term from your sum beforehand, $$S_{k+1}=\sum_{i=1}^{k+1}{k+1\choose i}(-1)^{i+1}\dfrac{1}{i}=\dfrac{(-1)^k}{k+1}+\sum_{i=1}^{k}{k+1\choose i}(-1)^{i+1}\dfrac{1}{i},$$ and then apply the identity to the second term: $$\sum_{i=1}^{k}{k+1\choose i}(-1)^{i+1}\dfrac{1}{i}=\sum_{i=1}^{k}{k\choose i}(-1)^{i+1}\dfrac{1}{i}+\sum_{i=1}^{k}{k\choose i-1}(-1)^{i+1}\dfrac{1}{i}$$ In brief, $$S_{k+1}=\dfrac{(-1)^k}{k+1}+S_k+\sum_{i=1}^{k}{k\choose i-1}(-1)^{i+1}\dfrac{1}{i}$$ Let's now fixate on computing the third term. Let's start by doing $i-1\to i$ leaving us with $$\sum_{i=1}^{k}{k\choose i-1}(-1)^{i+1}\dfrac{1}{i}=\sum_{i=0}^{k-1}{k\choose i}(-1)^{i}\dfrac{1}{i+1}$$ Moreover, $$\begin{aligned} \sum_{i=0}^{k-1}{k\choose i}(-1)^{i}\dfrac{1}{i+1}&=\int_0^1\sum_{i=0}^{k-1}{k\choose i}(-x)^{i}\,\mathrm dx\\ &=\int_0^1(1-x)^{k}-(-1)^kx^k\,\mathrm dx\\ &=-\dfrac{(-1)^k}{k+1}+\int_0^1(1-x)^{k}\,\mathrm dx \end{aligned}$$ Thus, $$S_{k+1}=S_k+\int_0^1(1-x)^{k}\,\mathrm dx=S_k+\dfrac{1}{k+1}$$ where we solved the integral by doing $1-x\to x$. Finally, by the inital hypothesis, $$S_k=\sum_{i=1}^{k}{k\choose i}(-1)^{i+1}\dfrac{1}{i}=\sum_{i=1}^k\dfrac{1}{i}$$ and hence $$S_{k+1}=\sum_{i=1}^k\dfrac{1}{i}+\dfrac{1}{k+1}=\sum_{i=1}^{k+1}\dfrac{1}{i}\qquad\blacksquare$$

Hug de Roda
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