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I'm stuck on this mathematical induction inequality and am not sure about what to do with it. Prove using the principle of mathematical induction that: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-....+(-1)^{n-1}\frac{1}{n}>0 \quad\forall\;n\in\mathbb Z^+$$ I'm not even sure how to even start it properly due to the signs alternating. I was thinking that I should work it out for $n=1$ and then $n=2$. But then when it reaches the inductive steps, I have no idea of what to do.

Thanks a lot!!

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  • $\begingroup$ Consider the cases of $n$ being odd or even seperately. In the even case note $1/(2n+1) \geq 1/(2n+2)$. $\endgroup$ – Donald Splutterwit Mar 20 '17 at 0:22
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O.K, first case $n=2$ which is $1-\frac{1}{2} >0 $ which is true.

Assume that its true for $n=2k$ for $k \geq 1$ which is $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} + \cdots + \frac{1}{2k-1}-\frac{1}{2k} >0$

We want to prove it for $n=2(k+1)$ which is $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} + \cdots + \frac{1}{2k-1}-\frac{1}{2k} +\frac{1}{2k+1}-\frac{1}{2k+2}>0$, by the assumption its true that the sum till $-\frac{1}{2k}$ is bigger than $0$ so we substitute by $0$, so it becomes $0+\frac{1}{2k-1}-\frac{1}{2k}>0$ which is $\frac{1}{2k-1} > \frac{1}{2k}$ which is clearly true. thus for all even cases the inequality holds true.(for odd cases we want that in order to get to even case we need to subtract some value and when reaching even case we know its bigger than $0$).

So if we have a value that we subtract from it some smaller value and we still bigger than $0$, so the initial values are certainly bigger than $0$.

Which conclude the proof.

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