Why Assume Limit of \( S_{n-1} = S_n \)?

[Mathematicsss]

Homework Statement

If the sum of a sub n to infinity (n=1) converges then the limit of n as n tends to infinity of an = 0

Homework Equations

The Attempt at a Solution

an =(a1+a2+...an)-(a1+...+an-1)
= limit of an (n tends to infinity) = sn -s(n-1) =0

The area I'm confused is why do we assume that the limit of s(n-1)= limit of sn

 

[fresh_42]

Homework Statement

If the sum of a sub n to infinity (n=1) converges then the limit of n as n tends to infinity of an = 0

Homework Equations

The Attempt at a Solution

an =(a1+a2+...an)-(a1+...+an-1)
= limit of an (n tends to infinity) = sn -s(n-1) =0

The area I'm confused is why do we assume that the limit of s(n-1)= limit of sn

The sum is given as convergent, so $\lim_{n \to \infty}S_n$ exists, let's say is equal to $S$. Can you simply try to prove, that $\lim_{n \to \infty} b_n = 0$ where $b_n = S_n - S_{n-1}\,$? Hint: try to estimate $b_n = (S_n-S) - (S_{n_1}-S)$.

 

[Mark44]

Homework Statement

If the sum of a sub n to infinity (n=1) converges then the limit of n as n tends to infinity of an = 0

Learning a little bit of LaTeX, will make what you're writing much easier to comprehend. Under INFO --> Help/How-to is this page: https://www.physicsforums.com/help/latexhelp/

Homework Equations

The Attempt at a Solution

an =(a1+a2+...an)-(a1+...+an-1)
= limit of an (n tends to infinity) = sn -s(n-1) =0

The area I'm confused is why do we assume that the limit of s(n-1)= limit of sn

How is $S_n$ defined? You don't mention it anywhere above.