Series: Leibniz’s test – sufficient but not necessary!?

Question

Let {a_n} be some sequence, a_n >=0.
Leibniz’s test is used to prove that alternating series
Σa_n*(-1)^n  is convergent.

It states that: if lim (a_n) =0 as n-> ∞ and the sequence {a_n}
is monotone decreasing then the alternating series Σa_n*(-1)^n  converges.

Conditions of the Leibniz’s test are sufficient but not necessary.

So, construct such alternating series which do not satisfy
conditions of the Leibniz’s test however converges and 
general term of which is a_n = f(n), where
1)	f(x) continuous;
2)	f(x) discontinuous,
x є R, n є N.

? · Accepted Answer

The condition lim (a_n) = 0 as n-> ∞ is necessary, so the only thing we can get rid of is monotonicity.

1. Let f(x) = sin^2(πx/2)/x^2.  Then f(x) ≥ 0 for all x.  For odd n, a_n = f(n) = 1/n^2, and for even n, a_n = f(n) = 0.  The sequence is not monotone decreasing, but this series is nonnegative and bounded above by the p-series with p = 2, so it converges.

2. Um, let f(x) = 1/x^2 when x is not an even integer, f(x) = 0 when x is an even integer.

I'm not sure why you distinguish between continuity and discontinuity of f(x)...  If a_n is a sequence of integers, it's easy to construct a continuous function f(x) so that f(n) = a_n: define f(x) to be piecewise linear, with f(x) = a_n(n+1-x)+a_(n+1)(x-n) for n ≤ x ≤ n+1.  And it's similarly easy to construct a discontinuous function f(x) so that f(n) = a_n: define f(x) to be some wacky discontinuous function for nonintegral x, and let f(n) = a_n.

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You're right, my example converges for reasons other than the fact it's alternating.  If you want something that's not absolutely convergent, consider:

a(n) = 1/n, n odd
a(n) = 1/(n+2), n even

Then the terms are not monotonically decreasing.  However, when n is odd, the partial sums are equal to:

-1-1/3+1/4-1/5+...-1/n+1/(n+1)

And when n is even, the partial sums are equal to:

-1-1/3+1/4-1/5+...-1/(n-1)+1/n+1/(n+2)

So if S(n) are the partial sums of this series and T(n) are the partial sums of the alternating harmonic series, we have:

S(n) = T(n+1)-1/2, n odd
S(n) = T(n+2)-1/2+1/(n+2), n even

Taking the limit as n goes to infinity, we see that this converges.

This is certainly not absolutely convergent, since it contains all terms of the harmonic series except 1/2.

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Series: Leibniz’s test – sufficient but not necessary!?

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