Hi,If I Have [math]\lim_{n \to \infty} a_n[/math] îs that the same with [math]\sum_{n=1}^{\infty} a_n[/math]?
I found that in my precalculus book but I Have no idea If this îs what it means.
If You Have a. Sum and after You reach the result why do You use the limit notation?
Do You add each term to reach a specific constant?
The limit of a sequence [imath]\displaystyle \lim_{n\rightarrow \infty}a_n[/imath] is different than the sum of a sequence [imath]\displaystyle \sum_{n=1}^{\infty}a_n[/imath], but they are related. If the limit converges, there is a possibility that the sum also converges. If the limit diverges, you can say directly the sum also diverges.
You can write the sum with a limit notation like this:
[imath]\displaystyle \sum_{n=1}^{\infty} a_n = \lim_{n\rightarrow \infty} \sum_{k=1}^{n} a_k[/imath]
Sometimes, you are given a sum:
[imath]\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n}[/imath]
And you can solve the sum by taking the limit of the partial sums. You can, for example calculate [imath]S_1, S_2, S_3, \ \text{and} \ S_4[/imath], and you will see a pattern for the partial sum, [imath]S_n[/imath], then you can calculate the limit of [imath]S_n[/imath] like this:
[imath]\displaystyle \lim_{n\rightarrow \infty} S_n = \lim_{n\rightarrow \infty} \left(1 - \frac{1}{2^n}\right) = 1 - 0 = 1[/imath]
Then you can claim that:
[imath]\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n} = 1[/imath]
