Defining summation of a function în a limit function

wolly

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Hi,If I Have limnan\lim_{n \to \infty} a_n îs that the same with n=1an\sum_{n=1}^{\infty} a_n?
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?
 
Hi,If I Have limnan\lim_{n \to \infty} a_n îs that the same with n=1an\sum_{n=1}^{\infty} a_n?
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?
Please give people a chance to respond. None of the Helpers take shifts to make sure there is 24/7 full coverage.

I'm not clear on what you are asking.

n=1an=limNn=1Nan\displaystyle \sum_{n = 1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n = 1}^N a_n

Does that answer your question?

-Dan
 
Please give people a chance to respond. None of the Helpers take shifts to make sure there is 24/7 full coverage.

I'm not clear on what you are asking.

n=1an=limNn=1Nan\displaystyle \sum_{n = 1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n = 1}^N a_n

Does that answer your question?

-Dan
I wanted to know why the summation of a function needs a limit to find the constant of that function.Are the sum and limit of that function the same? The answers to my questions is this expression which you gave me:
n=1an=limNn=1Nan\sum_{n = 1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n = 1}^N a_n?????
 
Hi,If I Have limnan\lim_{n \to \infty} a_n îs that the same with n=1an\sum_{n=1}^{\infty} a_n?
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 limnan\displaystyle \lim_{n\rightarrow \infty}a_n is different than the sum of a sequence n=1an\displaystyle \sum_{n=1}^{\infty}a_n, 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:

n=1an=limnk=1nak\displaystyle \sum_{n=1}^{\infty} a_n = \lim_{n\rightarrow \infty} \sum_{k=1}^{n} a_k

Sometimes, you are given a sum:

n=112n\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n}

And you can solve the sum by taking the limit of the partial sums. You can, for example calculate S1,S2,S3, and S4S_1, S_2, S_3, \ \text{and} \ S_4, and you will see a pattern for the partial sum, SnS_n, then you can calculate the limit of SnS_n like this:

limnSn=limn(112n)=10=1\displaystyle \lim_{n\rightarrow \infty} S_n = \lim_{n\rightarrow \infty} \left(1 - \frac{1}{2^n}\right) = 1 - 0 = 1

Then you can claim that:

n=112n=1\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n} = 1
 
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