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TIFR 2013 Problem 16 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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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