Need a bit of help with a summation proof

[daon]

I've been on this 3-part question for a long time. I've managed to geth through the first two but having trouble with the third part.

I have already proven the following:

Proven:

(1) \(\displaystyle \L \sum _{n=0}^{\infty} (x_n - x_{n+k}) = x_0+x_1+...+x_{k-1}-kx\)

(2) \(\displaystyle \L \sum _{n=0}^{\infty} \frac{1}{(n+a)(n+a+k)} = \frac{1}{k}[\frac{1}{a} + \frac{1}{a+1} + ... + \frac{1}{a+k-1}]\)

Need to prove: Given a natural number m,

\(\displaystyle \L \sum _{n=0}^{\infty} \frac{1}{(n + a)(n + a + m) ... (n + a + km)} = \frac{1}{km} \sum _{n=0}^{m-1} \frac{1}{(n + a)(n + a + m) ... (n + a + (k - 1)m)}\)

I have attempted to show this by induction on m. I have proven the basis when m=1, but I'm a mess trying to prove the inductive step. Should I induct on k?

 

[daon]

In case anyone cares, I found the solution. It didn't need induction afterall... now I feel like a doofus .

\(\displaystyle \L \sum \frac{1}{(n+a)(n+a+m)...(n+a+km)} = \frac{1}{km} \sum [ \frac{1}{(n+a)(n+a+m)(n+a+2m)...(n+a +(k-1)m)} - \frac{1}{(n+a+m)(n+a+2m)...(n + a + km)} ]\)
(to verify, combine fractions and notice that there is a km in the numerator).

Thus if we label:

\(\displaystyle \L x_n = \frac{1}{(n+a)(n+a+m)(n+a+2m)...(n+a +(k-1)m)}\)

Then

\(\displaystyle \L x_{(n+m)} = \frac{1}{(n+a+m)(n+a+2m)(n+a+3m)...(n+a +km)}\)

So, we have

\(\displaystyle \L \frac{1}{km}\sum [x_n - x_{n+m}]\)

And, by (1) in the original post, this equals:

\(\displaystyle \L \frac{1}{km}(x_0 + x_1 + ... + x_{m-1} - mx)\)

Where x by definition is the limit of \(\displaystyle x_n\) = 0.

So, we have

\(\displaystyle \L \frac{1}{km}(x_0 + x_1 + ... + x_{m-1}) = \frac{1}{km} \sum_{n=0}^{m-1} x_n\)

Substituting in \(\displaystyle x_n\) from above yeilds our desired result.

-Daon