convergence

**iLikeMaths** · January 22nd 2009, 11:11 PM

I have been having trouble with this question for days and i am having difficulty understanding what the question is asking me, its on the subject of series convergence and divergence, its a long question

This question concerns grouping the terms of a series as follows:
a0+(a1+a2)+(a3+a4)+(a5+· · ·+a8)+(a9+· · ·+a16)+(a17+· · ·+a32)+· · ·

Suppose that $0<a_{n+1}<a_{n}$ for all $n$ .

Show that
$\frac {1}{2}(2^{n+1} a_{2n+1})<a_{2n+1}+....<2^n a_{2n}$

hence show that if this series

$\sum a_n$ converges then this series $\sum 2^n a_2n$ also converges. and that if one of the two series diverges then so does the other.

hence determine if these series converge or diverge
$\sum \frac {1}{n \log n}$ and $\sum \frac {1} {n(\log n)^2}$

**Grandad** · January 23rd 2009, 01:02 AM

Hello iLikeMaths

Originally Posted by iLikeMaths

Show that
$\frac {1}{2}(2^{n+1} a_{2n+1})<a_{2n+1}+....<2^n a_{2n}$

Are you sure this shouldn't be:

$\frac {1}{2}(2^{n+1} a_{2^n+1})<a_{2^n+1}+...<2^n a_{2^n}$ where the expression in the middle (the one with the dots ...) actually means $a_{2^n+1}+... a_{2^{n+1}}$ (and not the sum to infinity which I originally took it to mean).

The reason I ask is that this ties in with the preamble to the question. For instance, look at the terms in the fourth set of brackets: $(a_9 + ... + a_{16})$

This corresponds to $n = 3$ , $2^n = 8$ , $2^n + 1 = 9$ . So it is the sequence

$a_{2^3+1} + ... + a_{2^4}$

There are $2^3 = 8$ terms in this bracket, each of which is less than the last term in the previous bracket, $a_8 (= a_{2^3})$ because $a_{n+1}<a_n$ for all $n$ , and so their sum is less than $2^3 \times a_{2^3}$

Generalise this to $n$ , and you get

$\sum_{i=2^n+1}^{2^{n+1}}a_i < 2^na_{2^n}$

which I think is what you want for the right-hand side of the inequality. The left-hand side is similar, but not quite the same.

To look at the convergence of $\sum a_n$ , I think you then use the fact that the sequence $a_{2^n+1}+... a_{2^{n+1}}$ is squeezed in between the other two.

Does that give you some ideas?

Grandad

**iLikeMaths** · January 23rd 2009, 01:33 AM

yes your post has been helpful, and i forgot to put the expression in the middle, could you please explain in more detail, because i am totally lost on the whole question, thank you

**Prove It** · January 23rd 2009, 02:36 AM

Originally Posted by iLikeMaths

I have been having trouble with this question for days and i am having difficulty understanding what the question is asking me, its on the subject of series convergence and divergence, its a long question

This question concerns grouping the terms of a series as follows:
a0+(a1+a2)+(a3+a4)+(a5+· · ·+a8)+(a9+· · ·+a16)+(a17+· · ·+a32)+· · ·

Suppose that $0<a_{n+1}<a_{n}$ for all $n$ .

Show that
$\frac {1}{2}(2^{n+1} a_{2n+1})<a_{2n+1}+....<2^n a_{2n}$

hence show that if this series

$\sum a_n$ converges then this series $\sum 2^n a_2n$ also converges. and that if one of the two series diverges then so does the other.

hence determine if these series converge or diverge
$\sum \frac {1}{n \log n}$ and $\sum \frac {1} {n(\log n)^2}$

I haven't studied sequences and series in quite a while, but couldn't you just use a ratio test to determine its convergence or divergence?

You're told $0 < a_{n+1} < a_n$ .

So $a_{n + 1} < a_n$ and thus $\frac{a_{n+1}}{a_n} < 1$ (we can divide and know it's not going to change the inequality because we're told both terms are positive).

Since $\frac{a_{n+1}}{a_n} < 1$ the series converges.

Can anyone see any flaws to my logic?

**Grandad** · January 23rd 2009, 02:50 AM

Hello Prove It

Originally Posted by Prove It

I haven't studied sequences and series in quite a while, but couldn't you just use a ratio test to determine its convergence or divergence?

You're told $0 < a_{n+1} < a_n$ .

So $a_{n + 1} < a_n$ and thus $\frac{a_{n+1}}{a_n} < 1$ (we can divide and know it's not going to change the inequality because we're told both terms are positive).

Since $\frac{a_{n+1}}{a_n} < 1$ the series converges.

Can anyone see any flaws to my logic?

Just a quick reply to this post. I think it has to be $\lim_{n \rightarrow \infty}| \frac{a_{n+1}}{a_n}| < 1$ for a convergent sequence, and we don't know this.

Haven't time to write more at present. Can anyone else take this up?

Grandad

**Grandad** · January 23rd 2009, 10:49 AM

Hello iLikeMaths

Here's a little more help.

To understand how this works, you must study carefully the way in which the brackets that group together the terms are formed. So, we group the sequence $\sum_{i=0}^\infty a_i$ as follows:

$a_0 + a_1 + a_2 + (a_3 + a_4) + (a_5 + ... + a_8) + (a_9 + ... + a_{16}) + (a_{17}+ ... +a_{32}) + ...$

In this grouping, then, after the first three terms $a_0$ , $a_1$ and $a_2$ , the number of terms in each bracket is $2, 4, 8, 16, ...$ , the first term in each bracket is $a_3, a_5, a_9, a_{17}, ...$ and the last term in each bracket is $a_4, a_8, a_{16}, a_{32}, ...$

Thus the $n^{th}$ bracket contains $2^n$ terms, the first term being $a_{2^n+1}$ and last term being $a_{2^{n+1}}$

Now this is a decreasing sequence of positive numbers; that is to say $0<a_{n+1}<a_n$ for all $n$ . So all the terms in the $n^{th}$ bracket are smaller than the last term in the $(n-1)^{th}$ bracket. And there are $2^n$ terms in the $n^{th}$ bracket. So:

$a_{2^n+1} + a_{2^n+2} + ... + a_{2^{n+1}} < a_{2^n} + a_{2^n} + ... + a_{2^n} = 2^na_{2^n}$

which is the RHS of the first inequality.

To get the LHS of this inequality, note that each term except the last in the $n^{th}$ bracket is greater than the last term in this bracket, which is $a_{2^{n+1}}$ . So:

$a_{2^n+1} + a_{2^n+2} + ... + a_{2^{n+1}} > a_{2^{n+1}} + a_{2^{n+1}} + ... + a_{2^{n+1}} = 2^na_{2^{n+1}}= \frac{1}{2}(2^{n+1}a_{2^{n+1}})$

So we have now shown that:

$\frac{1}{2}(2^{n+1}a_{2^{n+1}}) < a_{2^n+1} + a_{2^n+2} + ... + a_{2^{n+1}}<2^na_{2^n}$

So when we sum all the parts of the inequality, we get:

$\frac{1}{2}\sum_{n=1}^\infty(2^{n+1}a_{2^{n+1}}) < \sum_{n=3}^\infty a_n < \sum_{n=1}^\infty 2^na_{2^n}$

$\Rightarrow \frac{1}{2}\sum_{n=2}^\infty(2^{n}a_{2^{n}}) < \sum_{n=3}^\infty a_n < \sum_{n=1}^\infty 2^na_{2^n}$

Can you understand all this? Can you see how this leads to the statements about convergence and divergence?

Grandad

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