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Looked for a post regarding this specific proof by induction, and can't seem to find one... please see below for the proposition and my attempt. It seems I'm getting stuck when it comes to manipulating the fractions during the inductive step, but I'm now concerned that my approach is somehow flawed.

Prove that for all $n \ge 1$, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots + \frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots + \frac {1}{2n}$.

$Proof.$ We will prove this via mathematical induction.

Base Case. For $n=1$ we see that $\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{2(1)-1}-\frac{1}{2(1)}=1-\frac{1}{2}=\frac{1}{2}=\frac{1}{2(1)}=\frac{1}{2n}$.

Inductive Step. Assume our proposition holds for $n=k\ge 1$. Now observe that

$$\begin{align*}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots + \frac{1}{2k-1}-\frac{1}{2k}+\frac{1}{2(k+1)-1}-\frac{1}{2(k+1)}&=\\ \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\cdots + \frac {1}{2k} + \frac{1}{2(k+1)-1}-\frac{1}{2(k+1)} \end{align*}$$ Stuck here ^^^

Wanting to get to here: $\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\cdots + \frac {1}{2(k+1)}.$

ClownInTheMoon
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1 Answers1

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This doesnot need induction to prove!

$$1+{1\over 2}+{1\over 3}+\cdots+{1\over 2n}=1+{1\over 2}+{1\over 3}+\cdots+{1\over 2n}$$ Subtracting $2\left({1\over 2}+{1\over 4}+{1\over 6}+\cdots+{1\over 2n}\right)=1+{1\over 2}+{1\over 3}+\cdots+{1\over n}$ from both sides $$\implies 1-{1\over 2}+{1\over 3}+\cdots-{1\over 2n}={1\over n+1}+\cdots+{1\over 2n}$$

Qwerty
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