I proved it true for the base case but have no idea how to implement the assumption that it's true for n=k when trying to prove for n=k+1.

Am i right in saying there would be two cases i.e. k is odd or even.

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    $\begingroup$ Hello, is really your summand independent of $i$? Because currently the sum evaluates to $(-1)^{n-1}$, which is not always positive... $\endgroup$ – yo' Sep 20 '14 at 6:25

You must only prove two base cases, the odd and the even: Let's suppose you already know that $S_k = \sum_{i=1}^k \frac{(-1)^{k-1}}{k}$ is positive for all $n\geq k \geq 1$. Then, if $n$ is even, obviously $S_{n+1}$ will be positive. On the other hand, if $n$ is odd,$$S_{n+1} = S_n -\frac{1}{n+1} = S_{n-1} +\left(\frac{1}{n} - \frac{1}{n+1}\right).$$

Which from your base in the odd case is clearly positive.


Hint : $\frac 1k-\frac 1{k+1}$ is always greater than zero.

  • $\begingroup$ Why down voted. Anything wrong here $\endgroup$ – Jasser Sep 20 '14 at 6:23

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