The discussion centers on the convergence of the series defined by \( S_n \) and the reasoning behind the assumption that \( \lim_{n \to \infty} S_{n-1} = \lim_{n \to \infty} S_n \). It is established that if the series converges, then \( \lim_{n \to \infty} a_n = 0 \) must hold true, where \( a_n = S_n - S_{n-1} \). The participants emphasize the importance of defining \( S_n \) clearly to avoid confusion and suggest using limits to demonstrate the relationship between \( S_n \) and \( S_{n-1} \).
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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)##.Mathematicsss said: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
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/Mathematicsss said: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
How is ##S_n## defined? You don't mention it anywhere above.Mathematicsss said: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