Question

Error estimate of Alternating Series (Proof)?

If the series ∑(-1)^n*(a(n)) converges to S, then the nth term of the sequence of partial sums of an alternating series approaches the sum of the series as n approaches ∞. The difference between the exact sum and the sum of the first n finite terms corresponds to the error in approximating the convergent sum by the nth partial sum.
How Can I proof that
|S-Sn|=∑(-1)^n+1a(n) from n+1 to ∞<an+1

Answer 1

Consider an alternating series:

`sum_(n=0)^oo (-1)^na_n`

with `a_n >=0` and satisfying the conditions:

`(1) " " lim_(n->oo) a_n = 0`

`(2) " " a_n >= a_(n+1)`

Based on Leibniz theorem the series is convergent and we have:

`sum_(n=0)^oo (-1)^na_n = L`

Consider now the sequence of partial sums of odd order:

`s_(2k+1) = sum_(j=0)^(2k+1) (-1)^ja_j = (a_0-a_1) + (a_2-a_3) + ... +(a_(2k)-a_(2k+1))`

as, based on `(2)`, `a_(2j) >= a_(2j+1)`, all the terms in the last sum are positive and then `s_(2n+1)` is a monotone increasing sequence.

But for a monotone increasing convergent sequence the limit is the upper bound, so that:

`s_(2k+1) <= L`

Similarly we have that the partial sums of even order:

`s_(2k) = sum_(j=0)^(2k) (-1)^ja_j = a_0 - (a_1 -a_2) - ... - (a_(2k-1)-a_(2k))`

form a monotone decreasing sequence with:

`s_(2k) >= L`

Now consider:

`abs(L-s_n)`

For `n` odd we have that `s_n <= L` and `s_(n+1) >= L`, then:

`abs (L-s_n) = L-s_n <= s_(n+1)-s_n = a_(n+1)`

while for `n` even we have `s_n >= L` and `s_(n+1) <= L`, and again:

`abs (L-s_n) = s_n-L <= s_n - s_(n+1) = a_(n+1)`

In both cases:

`abs (L-s_n) <= a_(n+1)`