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
The proof of divergence test is a mathematical method used to determine if an infinite series diverges, meaning that its sum approaches infinity. It involves comparing the terms of a series to those of a known divergent series, such as the harmonic series, to determine whether the series itself also diverges.
The proof of divergence test works by using the Limit Comparison Test, which compares the ratio of the terms of the series in question to those of a known divergent series. If the limit of this ratio is greater than zero or infinity, then the series diverges. If the limit is zero or a finite number, then the series converges.
The proof of divergence test should be used when the terms of a series do not approach zero, or when the terms alternate between positive and negative values. It is also useful for series with polynomial or exponential terms.
No, the proof of divergence test can only be used to determine divergence. If the test shows that a series diverges, it is not conclusive proof and further tests must be performed to determine if the series converges or diverges.
Yes, the proof of divergence test is limited to series with non-negative terms and cannot be used on alternating series. It also cannot be used to determine the exact value of the sum of a series, only whether it converges or diverges.