Series Convergence: Understanding L'hopital's Rule | Simple Question Answered

[KAS90]

hey there..
I have this series for example:
∑_(n=1)^∞ a n-1 (2n-1/2n)If we apply L'hopital's rule to find the limit, and to see if the series is convergent or divergent, I think we will obtain the answer 1..
so does that mean the series converges towards one (1)? When can I decide a series is convergent or divergent? I'm really confused..

Appreciate ur help.. a lot..I really need to understand..

 

[HallsofIvy]

hey there..
I have this series for example:
∑_(n=1)^∞ a n-1 (2n-1/2n)If we apply L'hopital's rule to find the limit, and to see if the series is convergent or divergent, I think we will obtain the answer 1..
so does that mean the series converges towards one (1)? When can I decide a series is convergent or divergent? I'm really confused..

Appreciate ur help.. a lot..I really need to understand..

What are you applying L'hopital's rule to? Are the individual terms "a_n-1 (2n-1)/2n"? What is "a_n-1"?

If you are referring to the sequence (2n-1)/2n, alone, and you mean "a<sub>n-1[/sup]= (2n-1)/2n", you hardly have to apply L'hopital's rule to that. Dividing both numerator and denominator by n, we get (2- 1/n)/2 which goes to 1 as n goes to infinity.

In order that a series converges, the terms themselves must go to 0. If I am interpreting your question, so that there is no unknown "an-1" then this series does NOT converge because the sequence of terms does not go to 0.</sub>

 

[KAS90]

What are you applying L'hopital's rule to? Are the individual terms "a_n-1 (2n-1)/2n"? What is "a_n-1"?

If you are referring to the sequence (2n-1)/2n, alone, and you mean "a<sub>n-1[/sup]= (2n-1)/2n", you hardly have to apply L'hopital's rule to that. Dividing both numerator and denominator by n, we get (2- 1/n)/2 which goes to 1 as n goes to infinity.

In order that a series converges, the terms themselves must go to 0. If I am interpreting your question, so that there is no unknown "an-1" then this series does NOT converge because the sequence of terms does not go to 0.</sub>

_{oh ok.. well yeh, that part of (an-1) is a constant.. which doesn't really count in the series.. so what I understand now is that when we get zero for an answer only, the series is said to be convergent.. otherwise it diverges.. but why? I know I ask a lot, but I really want to understand the whole concept..}

 

[king vitamin]

oh ok.. well yeh, that part of (a_n-1) is a constant.. which doesn't really count in the series.. so what I understand now is that when we get zero for an answer only, the series is said to be convergent.. otherwise it diverges.. but why? I know I ask a lot, but I really want to understand the whole concept..

Just because the limit goes to zero does not imply convergence. The famous example is the series 1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ... where the terms to go zero, but adding up all of the terms goes to infinity. If the limit of the sequence being summed goes to zero, you need to use other tests for convergence (comparison test, integral test, ratio test) to determine whether the series converges.

In your original example (and any example where the limit of the sequence is finite), it's easy to see why the series diverges, because as n goes to infinity, the partial sums become closer to 1 + 1 + 1 + 1 + ... which will go to infinity.