# Another Proof by Induction

1. Sep 19, 2004

### Parth Dave

Show that the sum of the series from j = 1 up to 2m of (-1)^(j+1)*(1/j) is equal to the sum of the series from j=1 up to m of (1/(m+j)).

it works for m = 1.
After assuming m = k is true, and plugging in m = k+1 i got

sum of the series from j = 1 up to 2(k+1) of (-1)^(j+1)*(1/j) is equal to the sum of the series from j=1 up to k+1 of (1/(k+j+1)).

Now the big problem im having is with the right side. The left side is ok, i realize that it can be written as the left side of the induction hypothesis plus the next two terms. Whats really bugging me tho is the right side. I cant see any way to form the induction hypothesis out of it. This is really because of the m+j in the series.

Because if i want to write the sum of the series from j=1 up to k+1 of (1/(k+j+1)), but with the upper limit as k, i would write it as:

the sum of the series from j=1 up to k of (1/(k+j+1)) + 1/2(k+1) (since j = k+1)

is that correct? And if it is i cant generate the induction left hand side. Meaning i cant get rid of the summation. Am i on the right track? where did i make a mistake? Any help would be much appreciated.

2. Sep 20, 2004

### Ethereal

I think I managed to do it. I assume what you're trying to prove is the following:
$$\sum_{j=1}^{2m} \frac{(-1)^{j+1}}{j} = \sum_{j=1}^m \frac{1}{j+m}$$

When you substitute m=k+1, sooner or later you would get $$\sum_{j=1}^k \frac{1}{k+j}$$on the LHS.

Here's a hint: $$\sum_{j=1}^k \frac{1}{k+j} = (\sum_{j=1}^k \frac{1}{k+1+j})\ - \ \frac{1}{k+1} \ + \ \frac{1}{(k+1)+k}$$

LaTex sure took a long time to create.

Last edited by a moderator: Sep 20, 2004