# Real Analysis Help. Series Convergence

• Nov 20th 2008, 01:17 PM
megamet2000
Real Analysis Help. Series Convergence
i would love if someone could help me with 2 problems im having trouble with.

1) Let $\displaystyle a_1, a_2, a_3$, ... be a decreasing sequence of positive numbers. Show that $\displaystyle a_1+a_2+a_3+$... converges if and only if $\displaystyle a_1+2a_2+4a_4+8a_8+$... converges. i saw a similar one on here a couple days ago but this is slightly different

2) Show that a power series $\displaystyle \sum\limits_{n = 1}^\infty {c_nx^n }\$ has the same radius of convergence as $\displaystyle \sum\limits_{n = 1}^\infty {c_{n+m}x^n }\$, for any positive integer m

thanks
• Nov 20th 2008, 02:29 PM
Plato
Here is the problem: $\displaystyle a_n > a_{n + 1} > 0\,,\,\sum\limits_{n = 1}^\infty {a_n } \,\& \,\sum\limits_{n = 0}^\infty {2^n a_{2^n } }$.

Using that $\displaystyle a_n$ is decreasing, then following shows it in one direction.
$\displaystyle \begin{gathered} a_1 + \underbrace {a_2 + a_3 }_{} + \underbrace {a_4 + a_5 + a_6 + a_7 }_{} + \underbrace {a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} + a_{15} }_{} + \cdots \hfill \\ \leqslant a_1 + 2a_2 + 4a_4 + 8a_8 \cdots \hfill \\ \end{gathered}$

Now note that:
$\displaystyle \begin{gathered} a_2 + \underbrace {a_3 + a_4 }_{} + \underbrace {a_5 + a_6 + a_7 + a_8 }_{} + \underbrace {a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} + a_{15} + a_{16} }_{} + \cdots \hfill \\ \geqslant a_2 + 2a_4 + 4a_8 + 8a_{16}\cdots \hfill \\ \end{gathered}$
Also note that if $\displaystyle \sum\limits_{n = 1}^\infty {a_n }$ converges that $\displaystyle \sum\limits_{n = 2}^\infty {2a_n }$ converges.

Can you finish?
• Nov 20th 2008, 04:47 PM
megamet2000
Quote:

Originally Posted by Plato
Here is the problem: $\displaystyle a_n > a_{n + 1} > 0\,,\,\sum\limits_{n = 1}^\infty {a_n } \,\& \,\sum\limits_{n = 0}^\infty {2^n a_{2^n } }$.

Using that $\displaystyle a_n$ is decreasing, then following shows it in one direction.
$\displaystyle \begin{gathered} a_1 + \underbrace {a_2 + a_3 }_{} + \underbrace {a_4 + a_5 + a_6 + a_7 }_{} + \underbrace {a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} + a_{15} }_{} + \cdots \hfill \\ \leqslant a_1 + 2a_2 + 4a_4 + 8a_8 \cdots \hfill \\ \end{gathered}$

Now note that:
$\displaystyle \begin{gathered} a_2 + \underbrace {a_3 + a_4 }_{} + \underbrace {a_5 + a_6 + a_7 + a_8 }_{} + \underbrace {a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} + a_{15} + a_{16} }_{} + \cdots \hfill \\ \geqslant a_2 + 2a_4 + 4a_8 + 8a_{16}\cdots \hfill \\ \end{gathered}$
Also note that if $\displaystyle \sum\limits_{n = 1}^\infty {a_n }$ converges that $\displaystyle \sum\limits_{n = 2}^\infty {2a_n }$ converges.

Can you finish?

thanks for the help i appreciate it. proofs have never been my strong point would u mind helping me finish this?

also can anyone help me with the second question?
• Nov 20th 2008, 05:25 PM
Mathstud28
Quote:

Originally Posted by megamet2000

2) Show that a power series $\displaystyle \sum\limits_{n = 1}^\infty {c_nx^n }\$ has the same radius of convergence as $\displaystyle \sum\limits_{n = 1}^\infty {c_{n+m}x^n }\$, for any positive integer m

thanks

http://www.mathhelpforum.com/math-he...er-series.html
• Nov 20th 2008, 06:05 PM
megamet2000
awesome thank you very much
• Nov 23rd 2008, 10:23 AM
megamet2000
does anyone have any tips on how to finish number 1?
• Nov 23rd 2008, 10:27 AM
Mathstud28
Quote:

Originally Posted by megamet2000
does anyone have any tips on how to finish number 1?

What are you having problems with? Plato practicially gave you the solution.
• Nov 23rd 2008, 11:36 AM
megamet2000
i dont understand the purpose of the braces. i get how $\displaystyle a_n<2^na_{2^n}$ but after he says "now note that" i dont get how that sequence is now greater than the last sequence
• Nov 23rd 2008, 12:27 PM
Plato
The first inequality tell us that $\displaystyle \sum\limits_{k = 1}^\infty {a_k } \leqslant \sum\limits_{k = 0}^\infty {2^k a_{2^k } }$.
So if $\displaystyle \sum\limits_{k = 0}^\infty {2^k a_{2^k } }$ converges then $\displaystyle \sum\limits_{k = 1}^\infty {a_k }$ converges.

Likewise, the second inequality tells us $\displaystyle \sum\limits_{k = 1}^\infty {2^{k - 1} a_{2^k } } \leqslant \sum\limits_{k = 2}^\infty {a_k }$.
By comparison both converge.