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Seydlitz

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## Homework Statement

Problem taken from Boas Mathematical Methods book, Section 14 page 35.

Prove that if ##S=\sum_{n=1}^{\infty} a_n## is an alternating series with ##|a_{n+1}|<|a_n|##, and ##\lim_{n \to \infty} a_n=0##, then ##|S-(a_1+a_2+...+a_n)|\leq|a_{n+1}|##.

## The Attempt at a Solution

Hint: Group the terms in the error as ##(a_{n+1}+a_{n+2})+(a_{n+3}+a_{n+4})+...## to show that the error has the same sign as ##a_{n+1}##

Because alternating series, and ##|a_{n+1}|<|a_n|##

If ##a_{n+1}>0## and ##a_{n+2}<0##, then ##(a_{n+1}+a_{n+2})>0##. Then the same thing will happen with the subsequent group for all ##n##, and the whole error will have the same sign as ##a_{n+1}##. The same can be shown if ##a_{n+1}<0##.

Hint: Group them (the error) as ##a_{n+1}+(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+...##

To show that the error has magnitude less than ##|a_{n+1}|##.

It is true that ##(a_{n+2}+a_{n+3})=(S_{n+3}-S_{n+1})##. It can be shown using induction that the group of the terms can be represented as the partial sum of the whole series as shown above for all##n##.

Hence it is simple to see that because the series is convergent the partial sum will tend to ##S## and the whole group of terms beside ##a_{n+1}## will go to 0. This implies the equation in the theorem.

Alternatively, it can be shown that because of alternating series ##(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+(...)<0## because of the grouping of the term. The positive odd terms will be smaller than the negative even terms after ##a_{n+1}## and from this we can directly also get the inequality. ##|S-S_n|\leq|a_{n+1}|##

Can you guys help me to verify my flow of thoughts and proof?

Thank You

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