# Quick question about a telescoping series

1. Feb 27, 2005

### Townsend

$$\sum_{j=k}^{\infty}\left\{{\frac{1}{b_{j}}-\frac{1}{b_{j+1}}}\right\}=\frac{1}{b_{k}}$$

This series holds for all real monotone sequences of $$b_j$$.

So if I were to carry this series out to say n I end up with a partial sum that looks like:

$$S_n=\frac{1}{b_k}-\frac{1}{b_{k+(n+1)}}$$

Now as n goes to infinity we are left with just $$b_k$$. This of course implies that $$\frac{1}{b_{k+(n+1)}}$$ goes to zero as n goes to infinity. So does this mean that the monotone sequence $$b_j$$ must equal {1,2,3,4,5,...,j} ? If not what exactly are the constraints on $$b_j$$ to make that series an identity?

Thanks for the help everyone.

JTB

2. Feb 27, 2005

### Townsend

Never mind...I figured it out. As long as $$b_j$$ is strictly increasing then the identity holds. The amount of jump between any two terms is irrelevant.

If there are any further comments please feel free to make them other wise I will let this thread die peacefully.

JTB

3. Feb 28, 2005

### NateTG

Actually, it's a bit cleaner to talk about it as:
$$\sum_{i=k}^{\infty} \left(a_i-a_{i+1}\right)$$
Then any partial sum can easily be evaluated:
$$\sum_{i=k}^{n} \left(a_i-a_{i+1}\right) = a_k-a_{n+1}$$
so we have
$$\lim_{n \rightarrow \infty} \sum_{i=k}^{n} \left(a_i-a_{i+1}\right) = \lim_{n \rightarrow \infty} a_k-a_{n+1}=a_k - \lim_{n \rightarrow \infty} a_{n+1}$$

There's no need to restrict the series to being monotone or real.