Quick question about a telescoping series

[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

 

[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

 

[NateTG]

\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.

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.