Induction with sum on left and right side.

[Tremoi]

Homework Statement

Does
\sum_{k=1}^{n}(\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{k=1}^{n}\frac{1}{k + n}
hold for 1 \leq n

Homework Equations

The Attempt at a Solution

It holds for n = 1. I assume that it should be done with induction but I can't find a way actually compare the two sums to each other. I then had an idea about maybe putting each sum on a common denominator and prove that the both denominators and the both numerators are equal but that's not true so I don't really know where to start.
The left terms can be rewritten as \frac{1}{2k(2k-1)} as well but that haven't really helped me either.

 

[praharmitra]

You can prove this using induction. Using the usual induction techniques, define f(n) = \sum\limits_{k=1}^{k=n}\left(\frac{1}{2k-1} - \frac{1}{2k}\right). You want to show f(n) = \sum\limits_{k=1}^{k=n} \frac{1}{k+n}. You've already shown that this is true for n=1. Assuming the statement is true for n
<br /> f(n+1)\\<br /> = \sum\limits_{k=1}^{k=n+1}\left(\frac{1}{2k-1} - \frac{1}{2k}\right) = f(n) + \frac{1}{2n+1} - \frac{1}{2n+2} \\<br /> = \sum\limits_{k=1}^{k=n} \frac{1}{k+n}+ \frac{1}{2n+1} - \frac{1}{2n+2}<br />
Now changing the variable in the sum k \to k&#039; = k-1. The sum becomes
<br /> f(n+1) \\<br /> = \sum\limits_{k&#039;=0}^{k&#039;=n-1} \frac{1}{k&#039;+n+1}+ \frac{1}{2n+1} - \frac{1}{2n+2} \\<br /> = \sum\limits_{k&#039;=1}^{k&#039;=n-1} \frac{1}{k&#039;+n+1} + \frac{1}{n+1}+ \frac{1}{2n+1} - \frac{1}{2n+2}\\<br /> = \sum\limits_{k&#039;=1}^{k&#039;=n-1} \frac{1}{k&#039;+n+1}+ \frac{1}{2n+1} + \frac{1}{2n+2}\\<br /> =\sum\limits_{k&#039;=1}^{k&#039;=n+1} \frac{1}{k&#039;+n+1}<br />

Thus, the statement is true!