Question

Prove that absolute convergence implies convergence converse is not true

Answer 1

Answer:

Theorem: Absolute Convergence implies Convergence

Theorem: Absolute Convergence implies ConvergenceHence the sequence of regular partial sums {Sn} is Cauchy and therefore must converge (compare this proof with the Cauchy Criterion for Series). The converse is not true because the series converges, but the corresponding series of absolute values does not converge.

Answer 2

To find,

Prove that absolute convergence implies convergence converse is not true.

Solution,

Suppose ∑^∞ _n=1 a$_n$  is absolutely convergent. Let

T$_j$ = |a$_1$| + |a$_2$| + ... + |a$_j$|

be the sequence of partial sums of absolute values, and

S$_j$ = a$_1$ + a$_2$ + ... + a$_j$

be the "regular" sequence of partial sums. Since the series converges absolutely, there exists an integer N such that:

| T$_n$ - T$_m$| = |a$_n$| + |a$_{n-1}$| + ... + |a$_{m+2}$| + |a$_{m+1}$| < ∞

if n > m > N. But we have by the triangle inequality that

| S$_n$ - S$_{ml}$ = | a$_n$ + a$_{n-1}$ + ... a$_{m+1}$ | ≤ |a$_n$| + |a$_{n-1}$| + ... |a$_{m+1}$| = | T$_n$ - T$_m$ | < ∞

Hence the sequence of regular partial sums {S$_n$} is Cauchy and therefore must converge (compare this proof with the Cauchy Criterion for Series).

The converse is not true because the series ∑^∞ _n=1  $\frac{-(1)^{n+1}}{n}$ converges, but the corresponding series of absolute values ∑^∞ _n=1 1/n do not converge.

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