How do you prove by induction that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^n}{n}$ is always positive.

The base case works out and the inductive step is n=k, so $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^k}{k}$ holds true for the statement. I am stuck on the n = k+1.

if any one can give an idea, it would help alot.

  • 6
    $\begingroup$ $ \frac{1}{2n-1}- \frac{1}{2n} > 0$ $\endgroup$ – Donald Splutterwit Aug 27 '17 at 20:02
  • 1
    $\begingroup$ HINT: separate even from odd $n$ $\endgroup$ – Bram28 Aug 27 '17 at 20:03
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    $\begingroup$ I suspect the final term in your sum should be $\cdots -\dfrac{(-1)^{n}}{n}$ or $\cdots +\dfrac{(-1)^{n+1}}{n}$ $\endgroup$ – Henry Aug 27 '17 at 20:14


  • Call the sum $S_n$. You want to show $S_n \gt 0$ for all positive integers $n$

  • Show it is true for $n=1$ and $n=2$

  • Show that if it is true for $n=2k$ then it is true for $n=2k+1$ and $n=2k+2$

  • Use induction


Hint: you might find it helpful to divide it into the cases even/uneven.

For the case that $n+1$ is uneven it is also useful to note that $\frac{1}{n}>\frac{1}{n+1}$ for all $n\in\mathbb N$.


Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^n}{n}$ is always positive is equivalent to proving that:

$1+\frac{1}{3}+\dots +\frac{1}{2n-1} > \frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2n}$

For all $n\geq 1$

Clearly this is true when $n=1$. Assuming this is true for $n=k$, how could you show this is true for $n=k+1$?


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