Answer to Question #195941 in Real Analysis for Saurabh Sharma

Question

Let Pun be a series of arbitrary terms. For all n ∈ N, let pn =

1

2

(un + |un|) and qn =

1

2

(un − |un|). Show

that

(a) If Pun is absolutely convergent, then both Ppn and Pqn are convergent.

(b) If Pun is conditionally convergent, then both Ppn and Pqn are divergent.

Answer 1

$P_{u_n} = \sum_{n=1}^{\infty} u_n \\ p_n = u_n + |u_n| \\ q_n = u_n - |u_n|$

a) Suppose $P_{u_n} = \sum_{n=1}^{\infty} u_n \\$ is absolutely convergent. This implies that $\sum_{n=1}^{\infty} |u_n|$ is convergent and hence $\sum_{n=1}^{\infty} u_n$ is also convergent (by the convergence of $\sum_{n=1}^{\infty} |u_n|$ )

Consider,

$P_{p_n}= \sum_{n=1}^{\infty} p_n = \sum_{n=1}^{\infty} (u_n + |u_n| )= \sum_{n=1}^{\infty} u_n + \sum_{n=1}^{\infty} |u_n|$

And since the sum of two convergent series is convergent, we can safely conclude that $P_{p_n}$ is convergent as required.

Also, consider

$P_{q_n} = \sum_{n=1}^{\infty} q_n = \sum_{n=1}^{\infty} (u_n - |u_n| )= \sum_{n=1}^{\infty} u_n -\sum_{n=1}^{\infty} |u_n|$

And since the difference of two convergent series is convergent, we can safely conclude that $P_{q_n}$ is convergent as required.

b) Suppose $P_{u_n} = \sum_{n=1}^{\infty} u_n \\$ is conditionally convergent. This implies that $\sum_{n=1}^{\infty} |u_n|$ is divergent and $\sum_{n=1}^{\infty} u_n$ is convergent.

Consider,

$P_{p_n} = \sum_{n=1}^{\infty} p_n = \sum_{n=1}^{\infty} (u_n + |u_n| )= \sum_{n=1}^{\infty} u_n + \sum_{n=1}^{\infty} |u_n|$

Since, the sum of a convergent series and a divergent series yields a divergent series, we can safely conclude that $P_{p_n}$ is a divergent series as required.

Also, consider

$P_{q_n} = \sum_{n=1}^{\infty} q_n = \sum_{n=1}^{\infty} (u_n -|u_n| )= \sum_{n=1}^{\infty} u_n -\sum_{n=1}^{\infty} |u_n|$

Since, the difference between a convergent series and a divergent series yields a divergent series, we can safely conclude that $P_{q_n}$ is a divergent series as required.