why it converges

[missdydon]

consider S(a,b) = ∑_(k=0)^∞ 1/((k+a)(k+b))

a) show that S(a,b) converges if a and b are both positive. Why cant either be negative?



[FONT=arial, helvetica, clean, sans-serif]ok i think it will converge to 0 since the denominator gets bigger as the positive numbers increase.
i tried to prove it was true by using telescoping/partial fractions and my answer was that it converged to 1.
i do not know if that is right or not.

do i use telescoping to prove this?

thank you[/FONT]

 

[daon2]

consider S(a,b) = ∑_(k=0)^∞ 1/((k+a)(k+b))

a) show that S(a,b) converges if a and b are both positive. Why cant either be negative?



[FONT=arial, helvetica, clean, sans-serif]ok i think it will converge to 0 since the denominator gets bigger as the positive numbers increase.
i tried to prove it was true by using telescoping/partial fractions and my answer was that it converged to 1.
i do not know if that is right or not.

do i use telescoping to prove this?

thank you[/FONT]

It certainly does not converge to 1 in general and ESPECIALLY not 0. Your first term (and subsequent terms) is positive if a and b are non-negative. In this case S(a,b) > 1/(ab) >0.

This will converge by comparison to a p-series. I will let you think of how to determine the series to compare it to.

 

[HallsofIvy]

I'm not clear on what the question really is but one obvious point: if a or b is a negative integer then at least one term of the series will not exist.

Also the individual numbers in the series go to 0 (that is a necessary condition that a series converge) but the series, being a sum of positive numbers, can't be 0.

 

[girodd]

I agree that we don't know what the question really is, but if neither a nor b is an integer, the answer has to be that the series converges (absolutely, in fact), as indicated by the first post, regardless of the sign of a and b. Since the expected result is that the series not converge for either a or b negative, it must be the case that a and b are specified to be integers and we were not told this. In this case, as the second post observes, one or more terms of the series is undefined.