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## Problem:True/False

Suppose \(\left \{a_i\right\}\) is a sequence in \(\mathbb{R}\) such that \(\sum|a_i||x_i|<\infty\) whenever \(\sum|x_i|<\infty\). Then \(\left \{a_i\right\}\) is a bounded sequence.

*Hint:*

For any \(r\in(0,1)\), \(\sum r^n <\infty \).

Also, if the radius of convergence of a power series is R, then R is given by \(limsup|a_n|^{1/n}=\frac{1}{R} \)

**Discussion:**

Of course, \(\sum|a_n||r|^{n}<\infty\) for any \(r\in(-1,1)\).

Recall that, \(\sum|a_n|x^{n}<\infty\) for \(|x|<m\) means that radius of convergence of the power series is atleast m.

If the radius of convergence of \(\sum|a_n|x^{n}\) is R then \(R\ge1\).

i.e, $$ limsup|a_n|^{1/n} = \frac{1}{R} \le 1 $$

Hence, there exists \(N\in \mathbb{N}\) such that \(sup\{|a_n|^{1/n} : n \ge k\} \le 1 \) for all \(k \ge N\) (This is from the definition of limsup of a sequence).

Hence, \(sup\{|a_n|^{1/n} : n \ge N\} \le 1 \). Therefore, for each \(n \ge N\) we have \(|a_n|^{1/n} \le 1 \) for all \(n \ge N\). (Because sup is supremum which is least *upper bound*).

A real number which is in between 0 and 1 when raised to any power stays in between 0 and 1.

This allows us to state that \(|a_n| \le 1\) for all \(n \ge N\).

There are only finitely many terms left in the sequence which may not bounded by 1. But taking the maximum of their absolute value and 1 together we get a bound for the whole sequence.

For any \(n\in \mathbb{N}\),

$$ |a_n| \le max\{1,|a_1|,|a_2|,…,|a_{N-1}| \} $$

## Helpdesk

**What is this topic:**Real Analysis**What are some of the associated concept:**Bounded sequence, limsup, Power series**Book Suggestions:**Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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