Well, I’m going to give a brief explanation, sorry I don’t have a real example on me. If the series follows the telescoping series rule, it is convergent. If it doesn’t, then it can be either divergent or convergent.
Pretty much, you start plugging the first couple of values for the infinte sum. If it is a telescoping series, you should notice a pattern where things are canceling. After you prove that they’re canceling, work the last couple of values of the sum (like “infinity - 1″, infinity {it’s acutally better practice to write this as the limit goes to infinity}. After you proof the cancelation again, take the remaining terms and add them together. This will give you a value of what the series is convergent to.
Pseudo-Ex: (Where each parathesis is a term after plugging in a value into the function and t goes to infinity)
f(x) = (1/n) - (1/(n+1))
(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5)….
(1/(t-2) - 1/(t-1)) + (1/(t-1) - (1/t)) + (1/t + 1/(t+1))
As you can see in the example, the begging terms cancel, the -1/2 is canceled by adding the positive 1/2 in the next term. So, everything in this example cancels, except for the 1 (first term) and the 1/(t+1) (last term). With t approaching infinity, 1/(t+1) = 0 so this series is convergent to 1.